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I have migrated these answers from @alpoge's comments (first comment), and marked it community wiki to avoid reputation. I have done the transcription mostly by hand (cut-and-pasting with TeX code is tough, or at least I don't know how to do it), and have edited lightly, so I have probably introduced errors.

Hey, so let's say $p > 5$ or something. When $t = 1$ this is a standard mixed character sum and can be bounded using Weil's bound (see his paper On Some Exponential Sums, the last equation on pp. 206)  . Also, unless $\chi$ has conductor $q$, the sum is $0$ (split the sum into fibres modulo $p^{t − 1}$ and you're summing a nontrivial additive character).

Otherwise, write $h \mathbin{=:} x + p k$ with $0 \le x < p$ and $k \in \mathbb Z/p^{t - 1}\mathbb Z$ to obtain $$ \sum_x e_q(x)\prod_i \chi(x - a_i)\overline\chi(x - b_i)\sum_k e_q\Bigl(C p k - A_\chi\sum_{1 \le j \ll t} \frac{(-1)^{j - 1}p^j k^j}j\sum_i (x - a_i)^{-j} - (x - b_i)^{-j}\Bigr), $$ where $A_\chi \in (\mathbb Z/p^{t - 1}\mathbb Z)^\times$. (Here I've written, via the $p$-adic logarithm (i.e., $(\mathbb Z/q\mathbb Z)^\times \cong \mathbb F_p^\times \times \mathbb Z/p^{t - 1}\mathbb Z$), $\chi(1 - p t) = e_q\bigl(-A_\chi\sum_{1 \le i \ll t} \frac{p^i t^i}i\bigr)$, i.e., as an additive character on $\mathbb Z/p^{t - 1}\mathbb Z$.)

Now, by writing $k \mathbin{=:} m + p^{t - 2}n$ with $0 \le m < p^{t - 2}$ and $n \in \mathbb Z/p\mathbb Z$ and examining the sum over $n$, one sees that, unless $$ A_\chi\sum_i (x - a_i)^{-1} - (x - b_i)^{-1} \equiv C \pmod{p^{t - 1}} $$ (which has $\mathrm O(1)$-many solutions for $x \in \mathbb F_p$), the inner sum is $0$. If $t = 2$ then we're done, and we get a bound of $\mathrm O(1)\mathbin\cdot p = \mathrm O(p)$, since there are $\mathrm O(1)$ $x$ remaining, and, for those $x$, the sum over $k$ is $p$.

Otherwise, when $t \ge 3$ and the linear term does vanish, instead write $k \mathrel{=:} m + p^{\lceil(t - 2)/2\rceil} n$ with $0 \le m < p^{\lceil(t - 2)/2\rceil}$ and $n \in \mathbb Z/p^{\lfloor(t - 2)/2\rfloor}\mathbb Z$, and observe that the sum becomes linear in $n$, with coefficient depending on $m$. Unless that coefficient vanishes, which I think only happens when $m = 0$ (or the $a_i$ and $b_i$ are in remarkable position, but let's ignore that possibility) (forgive me if I'm being sloppy–I figured I'd give a quick answer to this just to help a bit), the sum over $n$ is $0$.

I think that the occurrence of both $\lceil\rceil$ and $\lfloor\rfloor$ in the previous comment may be a mistake, but I'm not sure which was intended.

Otherwise, $x$ has $\mathrm O(1)$ choices, $m = 0$, and the sum over $n$ is $p^{\lfloor(t - 2)/2\rfloor}$, so it seems like you get a bound of $\ll p^{\lfloor(t - 2)/2\rfloor}$ when $t > 2$. Forgive me if I've made a mistake! Just let me know and I'll try to fix it; I admit I'm just writing what comes to mind immediately, so it could be total nonsense. (For example, it seems the exponent of $p$ should have a ceiling or something, since, for $t = 3$, one shouldn't get a bound of $\mathrm O(1)$.)

I have migrated these answers from @alpoge's comments (first comment), and marked it community wiki to avoid reputation. I have done the transcription mostly by hand (cut-and-pasting with TeX code is tough, or at least I don't know how to do it), and have edited lightly, so I have probably introduced errors.

Hey, so let's say $p > 5$ or something. When $t = 1$ this is a standard mixed character sum and can be bounded using Weil's bound (see Weil - On some exponential sums, the last equation on p. 206). Also, unless $\chi$ has conductor $q$, the sum is $0$ (split the sum into fibres modulo $p^{t − 1}$ and you're summing a nontrivial additive character).

Otherwise, write $h \mathbin{=:} x + p k$ with $0 \le x < p$ and $k \in \mathbb Z/p^{t - 1}\mathbb Z$ to obtain $$ \sum_x e_q(x)\prod_i \chi(x - a_i)\overline\chi(x - b_i)\sum_k e_q\Bigl(C p k - A_\chi\sum_{1 \le j \ll t} \frac{(-1)^{j - 1}p^j k^j}j\sum_i (x - a_i)^{-j} - (x - b_i)^{-j}\Bigr), $$ where $A_\chi \in (\mathbb Z/p^{t - 1}\mathbb Z)^\times$. (Here I've written, via the $p$-adic logarithm (i.e., $(\mathbb Z/q\mathbb Z)^\times \cong \mathbb F_p^\times \times \mathbb Z/p^{t - 1}\mathbb Z$), $\chi(1 - p t) = e_q\bigl(-A_\chi\sum_{1 \le i \ll t} \frac{p^i t^i}i\bigr)$, i.e., as an additive character on $\mathbb Z/p^{t - 1}\mathbb Z$.)

Now, by writing $k \mathbin{=:} m + p^{t - 2}n$ with $0 \le m < p^{t - 2}$ and $n \in \mathbb Z/p\mathbb Z$ and examining the sum over $n$, one sees that, unless $$ A_\chi\sum_i (x - a_i)^{-1} - (x - b_i)^{-1} \equiv C \pmod{p^{t - 1}} $$ (which has $\mathrm O(1)$-many solutions for $x \in \mathbb F_p$), the inner sum is $0$. If $t = 2$ then we're done, and we get a bound of $\mathrm O(1)\mathbin\cdot p = \mathrm O(p)$, since there are $\mathrm O(1)$ $x$ remaining, and, for those $x$, the sum over $k$ is $p$.

Otherwise, when $t \ge 3$ and the linear term does vanish, instead write $k \mathrel{=:} m + p^{\lceil(t - 2)/2\rceil} n$ with $0 \le m < p^{\lceil(t - 2)/2\rceil}$ and $n \in \mathbb Z/p^{\lfloor(t - 2)/2\rfloor}\mathbb Z$, and observe that the sum becomes linear in $n$, with coefficient depending on $m$. Unless that coefficient vanishes, which I think only happens when $m = 0$ (or the $a_i$ and $b_i$ are in remarkable position, but let's ignore that possibility) (forgive me if I'm being sloppy–I figured I'd give a quick answer to this just to help a bit), the sum over $n$ is $0$.

I think that the occurrence of both $\lceil\rceil$ and $\lfloor\rfloor$ in the previous comment may be a mistake, but I'm not sure which was intended.

Otherwise, $x$ has $\mathrm O(1)$ choices, $m = 0$, and the sum over $n$ is $p^{\lfloor(t - 2)/2\rfloor}$, so it seems like you get a bound of $\ll p^{\lfloor(t - 2)/2\rfloor}$ when $t > 2$. Forgive me if I've made a mistake! Just let me know and I'll try to fix it; I admit I'm just writing what comes to mind immediately, so it could be total nonsense. (For example, it seems the exponent of $p$ should have a ceiling or something, since, for $t = 3$, one shouldn't get a bound of $\mathrm O(1)$.)

I have migrated these answers from @alpoge's comments (first comment), and marked it community wiki to avoid reputation. I have done the transcription mostly by hand (cut-and-pasting with TeX code is tough, or at least I don't know how to do it), and have edited lightly, so I have probably introduced errors.

Hey, so let's say $p > 5$ or something. When $t = 1$ this is a standard mixed character sum and can be bounded using the usual techniques (e.g., see Katz's webpage and click any paper with "mixed character sums" in the title). Also, unless $\chi$ has conductor $q$, the sum is $0$ (split the sum into fibres modulo $p^{t − 1}$ and you're summing a nontrivial additive character).

Otherwise, write $h \mathbin{=:} x + p k$ with $0 \le x < p$ and $k \in \mathbb Z/p^{t - 1}\mathbb Z$ to obtain $$ \sum_x e_q(x)\prod_i \chi(x - a_i)\overline\chi(x - b_i)\sum_k e_q\Bigl(C p k - A_\chi\sum_{1 \le j \ll t} \frac{(-1)^{j - 1}p^j k^j}j\sum_i (x - a_i)^{-j} - (x - b_i)^{-j}\Bigr), $$ where $A_\chi \in (\mathbb Z/p^{t - 1}\mathbb Z)^\times$. (Here I've written, via the $p$-adic logarithm (i.e., $(\mathbb Z/q\mathbb Z)^\times \cong \mathbb F_p^\times \times \mathbb Z/p^{t - 1}\mathbb Z$), $\chi(1 - p t) = e_q\bigl(-A_\chi\sum_{1 \le i \ll t} \frac{p^i t^i}i\bigr)$, i.e., as an additive character on $\mathbb Z/p^{t - 1}\mathbb Z$.)

Now, by writing $k \mathbin{=:} m + p^{t - 2}n$ with $0 \le m < p^{t - 2}$ and $n \in \mathbb Z/p\mathbb Z$ and examining the sum over $n$, one sees that, unless $$ A_\chi\sum_i (x - a_i)^{-1} - (x - b_i)^{-1} \equiv C \pmod{p^{t - 1}} $$ (which has $\mathrm O(1)$-many solutions for $x \in \mathbb F_p$), the inner sum is $0$. If $t = 2$ then we're done, and we get a bound of $\mathrm O(1)\mathbin\cdot p = \mathrm O(p)$, since there are $\mathrm O(1)$ $x$ remaining, and, for those $x$, the sum over $k$ is $p$.

Otherwise, when $t \ge 3$ and the linear term does vanish, instead write $k \mathrel{=:} m + p^{\lceil(t - 2)/2\rceil} n$ with $0 \le m < p^{\lceil(t - 2)/2\rceil}$ and $n \in \mathbb Z/p^{\lfloor(t - 2)/2\rfloor}\mathbb Z$, and observe that the sum becomes linear in $n$, with coefficient depending on $m$. Unless that coefficient vanishes, which I think only happens when $m = 0$ (or the $a_i$ and $b_i$ are in remarkable position, but let's ignore that possibility) (forgive me if I'm being sloppy–I figured I'd give a quick answer to this just to help a bit), the sum over $n$ is $0$.

I think that the occurrence of both $\lceil\rceil$ and $\lfloor\rfloor$ in the previous comment may be a mistake, but I'm not sure which was intended.

Otherwise, $x$ has $\mathrm O(1)$ choices, $m = 0$, and the sum over $n$ is $p^{\lfloor(t - 2)/2\rfloor}$, so it seems like you get a bound of $\ll p^{\lfloor(t - 2)/2\rfloor}$ when $t > 2$. Forgive me if I've made a mistake! Just let me know and I'll try to fix it; I admit I'm just writing what comes to mind immediately, so it could be total nonsense. (For example, it seems the exponent of $p$ should have a ceiling or something, since, for $t = 3$, one shouldn't get a bound of $\mathrm O(1)$.)

I have migrated these answers from @alpoge's comments (first comment), and marked it community wiki to avoid reputation. I have done the transcription mostly by hand (cut-and-pasting with TeX code is tough, or at least I don't know how to do it), and have edited lightly, so I have probably introduced errors.

Hey, so let's say $p > 5$ or something. When $t = 1$ this is a standard mixed character sum and can be bounded using Weil's bound (see his paper On Some Exponential Sums, the last equation on pp. 206)  . Also, unless $\chi$ has conductor $q$, the sum is $0$ (split the sum into fibres modulo $p^{t − 1}$ and you're summing a nontrivial additive character).

Otherwise, write $h \mathbin{=:} x + p k$ with $0 \le x < p$ and $k \in \mathbb Z/p^{t - 1}\mathbb Z$ to obtain $$ \sum_x e_q(x)\prod_i \chi(x - a_i)\overline\chi(x - b_i)\sum_k e_q\Bigl(C p k - A_\chi\sum_{1 \le j \ll t} \frac{(-1)^{j - 1}p^j k^j}j\sum_i (x - a_i)^{-j} - (x - b_i)^{-j}\Bigr), $$ where $A_\chi \in (\mathbb Z/p^{t - 1}\mathbb Z)^\times$. (Here I've written, via the $p$-adic logarithm (i.e., $(\mathbb Z/q\mathbb Z)^\times \cong \mathbb F_p^\times \times \mathbb Z/p^{t - 1}\mathbb Z$), $\chi(1 - p t) = e_q\bigl(-A_\chi\sum_{1 \le i \ll t} \frac{p^i t^i}i\bigr)$, i.e., as an additive character on $\mathbb Z/p^{t - 1}\mathbb Z$.)

Now, by writing $k \mathbin{=:} m + p^{t - 2}n$ with $0 \le m < p^{t - 2}$ and $n \in \mathbb Z/p\mathbb Z$ and examining the sum over $n$, one sees that, unless $$ A_\chi\sum_i (x - a_i)^{-1} - (x - b_i)^{-1} \equiv C \pmod{p^{t - 1}} $$ (which has $\mathrm O(1)$-many solutions for $x \in \mathbb F_p$), the inner sum is $0$. If $t = 2$ then we're done, and we get a bound of $\mathrm O(1)\mathbin\cdot p = \mathrm O(p)$, since there are $\mathrm O(1)$ $x$ remaining, and, for those $x$, the sum over $k$ is $p$.

Otherwise, when $t \ge 3$ and the linear term does vanish, instead write $k \mathrel{=:} m + p^{\lceil(t - 2)/2\rceil} n$ with $0 \le m < p^{\lceil(t - 2)/2\rceil}$ and $n \in \mathbb Z/p^{\lfloor(t - 2)/2\rfloor}\mathbb Z$, and observe that the sum becomes linear in $n$, with coefficient depending on $m$. Unless that coefficient vanishes, which I think only happens when $m = 0$ (or the $a_i$ and $b_i$ are in remarkable position, but let's ignore that possibility) (forgive me if I'm being sloppy–I figured I'd give a quick answer to this just to help a bit), the sum over $n$ is $0$.

I think that the occurrence of both $\lceil\rceil$ and $\lfloor\rfloor$ in the previous comment may be a mistake, but I'm not sure which was intended.

Otherwise, $x$ has $\mathrm O(1)$ choices, $m = 0$, and the sum over $n$ is $p^{\lfloor(t - 2)/2\rfloor}$, so it seems like you get a bound of $\ll p^{\lfloor(t - 2)/2\rfloor}$ when $t > 2$. Forgive me if I've made a mistake! Just let me know and I'll try to fix it; I admit I'm just writing what comes to mind immediately, so it could be total nonsense. (For example, it seems the exponent of $p$ should have a ceiling or something, since, for $t = 3$, one shouldn't get a bound of $\mathrm O(1)$.)

I have migrated these answers from @alpoge's comments (first comment), and marked it community wiki to avoid reputation. I have done the transcription mostly by hand (cut-and-pasting with TeX code is tough, or at least I don't know how to do it), and have edited lightly, so I have probably introduced errors.

Hey, so let's say $p > 5$ or something. When $t = 1$ this is a standard mixed character sum and can be bounded using the usual techniques (e.g., see Katz's webpage and click any paper with "mixed character sums" in the title). Also, unless $\chi$ has conductor $q$, the sum is $0$ (split the sum into fibres modulo $p^{t − 1}$ and you're summing a nontrivial additive character).

Otherwise, write $h \mathbin{=:} x + p k$ with $0 \le x < p$ and $k \in \mathbb Z/p^{t - 1}\mathbb Z$ to obtain $$ \sum_x e_q(x)\prod_i \chi(x - a_i)\overline\chi(x - b_i)\sum_k e_q\Bigl(C p k - A_\chi\sum_{1 \le j \ll t} \frac{(-1)^{j - 1}p^j k^j}j\sum_i (x - a_i)^{-j} - (x - b_i)^{-j}\Bigr), $$ where $A_\chi \in (\mathbb Z/p^{t - 1}\mathbb Z)^\times$. (Here I've written, via the $p$-adic logarithm (i.e., $(\mathbb Z/q\mathbb Z)^\times \cong \mathbb F_p^\times \times \mathbb Z/p^{t - 1}\mathbb Z$), $\chi(1 - p t) = e_q\bigl(-A_\chi\sum_{1 \le i \ll t} \frac{p^i t^i}i\bigr)$, i.e., as an additive character on $\mathbb Z/p^{t - 1}\mathbb Z$.)

Now, by writing $k \mathbin{=:} m + p^{t - 2}n$ with $0 \le m < p^{t - 2}$ and $n \in \mathbb Z/p\mathbb Z$ and examining the sum over $n$, one sees that, unless $$ A_\chi\sum_i (x - a_i)^{-1} - (x - b_i)^{-1} \equiv C \pmod{p^{t - 1}} $$ (which has $\mathrm O(1)$-many solutions for $x \in \mathbb F_p$), the inner sum is $0$. If $t = 2$ then we're done, and we get a bound of $\mathrm O(1)\mathbin\cdot p = \mathrm O(p)$, since there are $\mathrm O(1)$ $x$ remaining, and, for those $x$, the sum over $k$ is $p$.

Otherwise, when $t \ge 3$ and the linear term does vanish, instead write $k \mathrel{=:} m + p^{\lceil(t - 2)/2\rceil} n$ with $0 \le m < p^{\lceil(t - 2)/2\rceil}$ and $n \in \mathbb Z/p^{\lfloor(t - 2)/2\rfloor}$, and observe that the sum becomes linear in $n$, with coefficient depending on $m$. Unless that coefficient vanishes, which I think only happens when $m = 0$ (or the $a_i$ and $b_i$ are in remarkable position, but let's ignore that possibility) (forgive me if I'm being sloppy–I figured I'd give a quick answer to this just to help a bit), the sum over $n$ is $0$.

I think that the occurrence of both $\lceil\rceil$ and $\lfloor\rfloor$ in the previous comment may be a mistake, but I'm not sure which was intended.

Otherwise, $x$ has $\mathrm O(1)$ choices, $m = 0$, and the sum over $n$ is $p^{\lfloor(t - 2)/2\rfloor}$, so it seems like you get a bound of $\ll p^{\lfloor(t - 2)/2\rfloor}$ when $t > 2$. Forgive me if I've made a mistake! Just let me know and I'll try to fix it; I admit I'm just writing what comes to mind immediately, so it could be total nonsense. (For example, it seems the exponent of $p$ should have a ceiling or something, since, for $t = 3$, one shouldn't get a bound of $\mathrm O(1)$.)

I have migrated these answers from @alpoge's comments (first comment), and marked it community wiki to avoid reputation. I have done the transcription mostly by hand (cut-and-pasting with TeX code is tough, or at least I don't know how to do it), and have edited lightly, so I have probably introduced errors.

Hey, so let's say $p > 5$ or something. When $t = 1$ this is a standard mixed character sum and can be bounded using the usual techniques (e.g., see Katz's webpage and click any paper with "mixed character sums" in the title). Also, unless $\chi$ has conductor $q$, the sum is $0$ (split the sum into fibres modulo $p^{t − 1}$ and you're summing a nontrivial additive character).

Otherwise, write $h \mathbin{=:} x + p k$ with $0 \le x < p$ and $k \in \mathbb Z/p^{t - 1}\mathbb Z$ to obtain $$ \sum_x e_q(x)\prod_i \chi(x - a_i)\overline\chi(x - b_i)\sum_k e_q\Bigl(C p k - A_\chi\sum_{1 \le j \ll t} \frac{(-1)^{j - 1}p^j k^j}j\sum_i (x - a_i)^{-j} - (x - b_i)^{-j}\Bigr), $$ where $A_\chi \in (\mathbb Z/p^{t - 1}\mathbb Z)^\times$. (Here I've written, via the $p$-adic logarithm (i.e., $(\mathbb Z/q\mathbb Z)^\times \cong \mathbb F_p^\times \times \mathbb Z/p^{t - 1}\mathbb Z$), $\chi(1 - p t) = e_q\bigl(-A_\chi\sum_{1 \le i \ll t} \frac{p^i t^i}i\bigr)$, i.e., as an additive character on $\mathbb Z/p^{t - 1}\mathbb Z$.)

Now, by writing $k \mathbin{=:} m + p^{t - 2}n$ with $0 \le m < p^{t - 2}$ and $n \in \mathbb Z/p\mathbb Z$ and examining the sum over $n$, one sees that, unless $$ A_\chi\sum_i (x - a_i)^{-1} - (x - b_i)^{-1} \equiv C \pmod{p^{t - 1}} $$ (which has $\mathrm O(1)$-many solutions for $x \in \mathbb F_p$), the inner sum is $0$. If $t = 2$ then we're done, and we get a bound of $\mathrm O(1)\mathbin\cdot p = \mathrm O(p)$, since there are $\mathrm O(1)$ $x$ remaining, and, for those $x$, the sum over $k$ is $p$.

Otherwise, when $t \ge 3$ and the linear term does vanish, instead write $k \mathrel{=:} m + p^{\lceil(t - 2)/2\rceil} n$ with $0 \le m < p^{\lceil(t - 2)/2\rceil}$ and $n \in \mathbb Z/p^{\lfloor(t - 2)/2\rfloor}\mathbb Z$, and observe that the sum becomes linear in $n$, with coefficient depending on $m$. Unless that coefficient vanishes, which I think only happens when $m = 0$ (or the $a_i$ and $b_i$ are in remarkable position, but let's ignore that possibility) (forgive me if I'm being sloppy–I figured I'd give a quick answer to this just to help a bit), the sum over $n$ is $0$.

I think that the occurrence of both $\lceil\rceil$ and $\lfloor\rfloor$ in the previous comment may be a mistake, but I'm not sure which was intended.

Otherwise, $x$ has $\mathrm O(1)$ choices, $m = 0$, and the sum over $n$ is $p^{\lfloor(t - 2)/2\rfloor}$, so it seems like you get a bound of $\ll p^{\lfloor(t - 2)/2\rfloor}$ when $t > 2$. Forgive me if I've made a mistake! Just let me know and I'll try to fix it; I admit I'm just writing what comes to mind immediately, so it could be total nonsense. (For example, it seems the exponent of $p$ should have a ceiling or something, since, for $t = 3$, one shouldn't get a bound of $\mathrm O(1)$.)