Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.11
Solution: Suppose $I \not\subseteq P$. Then there exists $a \in I$ such that $a \notin P$. Now let $b \in J$. Since $IJ \subseteq P$, $ab \in P$. Since $P$ is a prime ideal and $a \notin P$, we have $b \in P$. Since $b$ was arbitrary in $J$, $J \subseteq P$.