= Solution
{c}
\b[True, without a Noetherian hypothesis.] By definition, the <symbolic power> is the <contraction of an ideal>
$$
P^{(n)}=P^nR_P\cap R,
$$
where the intersection notation means inverse image under $R\to R_P$, even if this map is not <injective>. In the <local ring> $A=R_P$, put $\mathfrak m=PR_P$. Then $P^nR_P=\mathfrak m^n$, and $\sqrt{\mathfrak m^n}=\mathfrak m$: every element of $\mathfrak m$ has its $n$th power in $\mathfrak m^n$, whereas a <unit> cannot have a power in this proper <ideal>.
The <ideal> $\mathfrak m^n$ is $\mathfrak m$-primary. Indeed, if $uv\in\mathfrak m^n$ and $v\notin\mathfrak m$, then $v$ is a <unit>, so $u\in\mathfrak m^n$. Its <contraction of an ideal> is therefore $P$-primary: its <radical of an ideal> contracts to $P$, and the same implication applies to the images of any two elements of $R$. Thus $P^{(n)}$ is always $P$-primary, proving the reverse implication immediately when $P^n=P^{(n)}$.
For the other implication, suppose $P^n$ is $P$-primary. Membership in the contraction can be expressed by clearing denominators:
$$
r\in P^{(n)}\quad\Longleftrightarrow\quad
sr\in P^n\text{ for some }s\notin P.
$$
For completeness, if $r/1=a/t$ with $a\in P^n$ and $t\notin P$, equality of fractions gives $u(tr-a)=0$ for some $u\notin P$, so $(ut)r=ua\in P^n$; the converse follows by inverting $s$. Since $s\notin\sqrt{P^n}=P$, the <primary ideal> property forces $r\in P^n$. The inclusion $P^n\subseteq P^{(n)}$ always holds, giving
$$
\boxed{P^n\text{ is }P\text{-primary}\iff P^n=P^{(n)}.}
$$
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