Consider the following: $r_1,...,r_t$ are iid symmetric signs taking value $\pm1$, independent of $B\sim Binomial(p, q)$ with integer $p$ and $q=p^{-1.01}$.

**Question**: Consider $t$ as a non-decreasing function of $p$. For which such non-decreasing function of $p$ does there exist a constant $\lambda>0$ independent of $p$ such that the following holds:
$$E[\exp(\lambda B \sum_{u=1}^t \frac{r_u}{\sqrt t})] \le 1.05,$$
for $p$ large enough.

**Remark:**
Note that $\tilde Z = \sum_u\frac{r_u}{\sqrt t}$ has variance 1 and is very close to $N(0,1)$, for instance by Tusnady's inequality. If $\tilde Z$ were replaced by 1 inside the exponential, we could proceed as follows: using the exact formula for the MGF of the binomial, and using $(1+x/p)^p\le e^x$,
$$
E[\exp(\lambda B)=
(1-q + qe^\lambda)^p
\le (1 + qe^\lambda)^p \le \exp(pq e^\lambda) \le \exp(p^{-0.01} e^{\lambda}),
$$
so that for any constant $\lambda$ (say, $\lambda=1$), the RHS is bounded from above by 1.05 or any other constant strictly greater than 1. If $t$ is constant, not growing to $\infty$ simultaneously as $p$, then we can use this remark to answer the question.

Edit: **Another remark**: the previous argument shows
$$E[\exp(\lambda \tilde Z B)=
(1-q + qe^{\lambda\tilde Z})^p
\le (1 + qe^{\lambda\tilde Z})^p \le \exp(pq e^{\lambda\tilde Z}) \le \exp(p^{-0.01} e^{\lambda\tilde Z})
$$
and since $\tilde Z\le \sqrt t$ always holds,
$\sqrt t = \log(p^{0.009})\asymp \log p$ with $\lambda=0$ works. The harder question is with $t$ varying more rapidly than this.