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True, without a Noetherian hypothesis. By definition, the symbolic power is the contraction of an idealwhere 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 $\mathfrak{m}=P{R}_P$. Then $P^n{R}_P={\mathfrak{m}}^n$, 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: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