$p$ cannot have more than $p$ of any prime factor, so if $a$ contains all of the prime factors of $p$ then certainly $a^p$ and $a^q$ do, otherwise not.

Hi ; I Think The Other Number Must Divisible By $q$ Best Regard

Let $a=\prod_{i=1}^{r}{{q_i}^{\alpha_i}}$ and $p=\prod_{j=1}^{s}{{p_j}^{\beta_j}}$ Now $p$ divides $a^p$ or $a^q$ means $(p_j)_{j=1}^{s}$ is a subset of $(q_i)_{i=1}^{r}$.Also the exponent $\beta_i$ of any $p_i$ cannot exceed $p$ since for some $i$ if it exceeds $p$ then $p=\prod_{j=1}^{s}{{p_j}^{\beta_j}} \ge {p_i}^{\beta_i} \ge {p_i}^{p} \ge 2^p>p$ a contradiction. It readily follows that $p|a^p \Leftrightarrow p|a^q$. Such a bad problem.

We will show that if $p\mid a^p$ or $p\mid a^q$, then it forces $p\mid a^p$, and since $a^p\mid a^q$ it will also be $p\mid a^q$. The case $p=1$ being trivial, assume $p\geq 2$. Consider any prime $t$ dividing $p$, so it must also divide $a$. Let $u\geq 1$ be the exponent of $t$ in $p$ and $v\geq 1$ be the exponent of $t$ in $a$. But for any values of $u$ and $v$ we have $u< t^u \leq p \leq vp$, so $t^u \mid ( t^v)^p$. This means $p\mid a^p$.