Bezout's theorem guarantees the existence of integers $m$ and $n$ satisfying
with $0\lt m\lt q$ and $0\lt n\lt p$. (Note, $m$ and $n$ are strictly positive since $p$ and $q$ are distinct primes.) Letting $M=q-m$ and $N=p-n$ we also have
with $0\lt M\lt q$ and $0\lt N\lt p$. Note that $n,N\lt p$ implies $q$ is the largest prime divisor of both $qn$ and $qN$ since $p\lt q$. As for $m$ and $M=q-m$, they can't both be larger than $p$ since that would imply $q=m+M\gt2p$. Thus one of them, at least, is less than or equal to $p$ and hence $p$ is the largest prime divisor of either $pm$ or $pM$ (or both).
As an example, let $p=47$ and $q=89$. We have
which is not what we want, since $53$ is a prime larger than $47$, but we also have
which does give us what we want.
Remark: It seems likely (or at least plausible) that for "most" pairs $(p,q)$, $p$ is the largest prime divisor of both $pm$ and $pM$. In looking for an example where it isn't, I got lucky: $(47,89)$ was the first pair I thought to try, and it produced $m=53$.