The Bombieri-Vinogradov Theorem states that given $A>0$, there exists $A>0$ such that for $Q=\frac{\sqrt{x}}{\left(\log x\right)^{B}},$ we have $$\sum_{q\leq Q}\max_{y\leq x}\max_{\begin{array}{c} a\text{ mod q}\\ (a,q)=1 \end{array}}\left|\psi(y;q,a)-\frac{y}{\phi(q)}\right|\ll \frac{x}{(\log x)^A}.$$

I was wondering what happens if a restriction is put on the $q$ so that they are all divisible by some smaller integer $k$. Are there any non-trivial bounds on the average over $q$ divisible by $k$? Specifically, suppose that $k\leq Q^{1-\epsilon},$ and that there is no Siegel zero${}^{++}$ for any $\chi$ modulo $k$. Is it true that $$\sum_{\begin{array}{c} q\leq Q\\ k|q \end{array}}\max_{y\leq x}\max_{\begin{array}{c} a\text{ mod q}\\ (a,q)=1 \end{array}}\left|\psi(y;q,a)-\frac{y}{\phi(q)}\right|\ll_{\epsilon}\frac{1}{k} \frac{x}{(\log x)^A}.$$

Any references would be greatly appreciated.

Thanks for your help,

${}^{++}$ Edit: As mentioned by Terence Tao in the comments, there is an issue regarding the Siegel zeros of the characters modulo $k$. The original result I was asking about would give stronger bounds on the location of certain Siegel zeros, and for that reason it is out of reach.

I have added a brief heuristic for anyone who is interested:

Heuristic: Suppose that $k$ is a small power of $x$. If there is an exceptional zero $\beta$ for a quadratic character $\chi$ modulo $k$ , then for every $q$ such that $k|q,$ the induced character modulo $q,$ $\chi^{\star},$ will have the same exceptional zero, and so we expect that $$\psi(x;q,a)\approx\frac{x}{\phi(q)}-\frac{\chi^{\star}(a)}{\phi(q)}\frac{x^{\beta}}{\beta}+small$$ for each $q.$ This leads us to expect that $$\sum_{\begin{array}{c} q\leq Q\\ k|q \end{array}}\max_{y\leq x}\max_{a\text{ mod }q}\left|\psi\left(y;q,a\right)-\frac{y}{\phi(q)}\right|\approx x^{\beta}\sum_{\begin{array}{c} q\leq Q\\ k|q \end{array}}\frac{1}{\phi(q)}\approx\frac{x^{\beta}\log x}{k},$$ however, the desired upper bound is too strong, as $$x^{\beta}\log x\ll \frac{x}{\left(\log x\right)^{B}},$$ implies that for some $C>0$, $$\beta\leq1-C\frac{\log\log x}{\log x},$$ which is stronger than the long standing bound by Siegel $$\beta\leq1-C(\epsilon)x^{-\epsilon}.$$ In the above, we have used $x$ interchangeably with $k$ for the bounds since $k$ is a small power of $x$.