 Solution

If $Q\mid P{O}_L$, then $P{O}_L\subseteq Q$, so $P\subseteq Q\cap O_K$. The contraction is a nonzero prime ideal of $O_K$ containing the maximal ideal $P$, hence equals $P$. Conversely, if $P=Q\cap O_K$, every element of $P$ lies in $Q$, so the ideal it generates satisfies $P{O}_L\subseteq Q$, equivalently $Q\mid P{O}_L$.