= Solution
{c}
\b[False.] In the <polynomial ring> $R=k[x,y]$ over a <field>, take
$$
P=(x,y),\qquad I=(x,y^2).
$$
The <ideal> $P$ is a <maximal ideal>, with <quotient ring> $R/P\cong k$. Also
$$
R/I\cong k[y]/(y^2).
$$
In this last <ring>, an element with nonzero constant term is a <unit>, and an element with zero constant term is a multiple of $y$ and has square zero. Thus every <zero divisor> is a <nilpotent element>. This is equivalent to $I$ being a <primary ideal>: a zero product in the <quotient ring> with one nonzero factor forces the other factor to be nilpotent. Its <radical of an ideal> is $P$, so $I$ is $P$-primary.
However, $y\notin I$, so $I\ne P$. Moreover $x\in I$ while $x\notin P^2$, since every polynomial in $P^2$ has no term of total degree one. For every $n\geq2$, $P^n\subseteq P^2$, so $I\ne P^n$. Hence
$$
\boxed{(x,y^2)\text{ is }(x,y)\text{-primary but is not a power of }(x,y).}
$$
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