Residue class of prime powers

Question

Assume $c(n)$ an arithmetic function that is multiplicative and has period $q$. For a fixed prime number $p$ with $p\mid q$ write $A_d=\{m\in\mathbb{N}:p^m\equiv d\text{ (mod }q)\}$ for $d\in\{0,1,\dotsc,p-1\}$. Prove that \begin{equation} \sum_{m=0}^\infty c(p^m)p^{-ms}=\sum_{d=0}^{p-1}c(d)\sum_{m\in A_d}p^{-ms}. \end{equation} I got a problem with the upper bound of the first sum on the right being $p-1$ instead of $q-1$. The whole thing would be clear if we could always find $d\in\{0,1,\dotsc,p-1\}$ with $p^m\equiv d\text{ (mod }q)$, but as there are counterexamples (see answers below), this is the wrong way.

Answer 1

No. Take $q=4$, $m=1$ and $p=2$.

Or $q=6$, $m=1$, $p=3$.