= Solution
{c}
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 $n\geq1$. Equivalently, when $\sqrt Q=P$, the implication is $ab\in Q$, $b\notin P\Rightarrow a\in Q$.
\b[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 $a\in Q$. Therefore
$$
\boxed{Q_1\cap\cdots\cap Q_m\text{ is }P\text{-primary}\quad(m\geq1).}
$$
As usual, the statement about finite intersections excludes the empty intersection, which is the whole <ring> and is not a <primary ideal>.
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