 Solution

A $P$-primary ideal is a proper ideal $Q$ with radical of an ideal $\sqrt{Q}=P$, such that $ab\in Q$ and $a\notin Q$ imply $b^n\in Q$ for some $n\ge 1$. Equivalently, when $\sqrt{Q}=P$, the implication is $ab\in Q$, $b\notin P\Rightarrow a\in Q$.

True for a nonempty finite family. Let $Q=Q_1\cap \cdots \cap Q_m$, with each $Q_j$ a $P$-primary ideal. This intersection of ideals is proper. Its radical of an ideal is $P$: if $r\in P$, choose $n_j$ with $r^{n_j}\in Q_j$ and put $N={\max }_j{n}_j$. Then $r^N\in Q$. Conversely, $r^N\in Q$ implies $r\in \sqrt{Q_j}=P$.

If $ab\in Q$ and $b\notin P$, the primary ideal property gives $a\in Q_j$ for every $j$, so $a\in Q$. ThereforeAs usual, the statement about finite intersections excludes the empty intersection, which is the whole ring and is not a primary ideal.