For example in the ring $R=k[x,y]$, the ideal $I=(x,y^2)$ is $P$-primary, where $P = (x, y)$, but is not a power of $P$, so the answer to your question is negative.
Let $S\subset\mathbb{Q}[x]$ be the subring consisting of polynomials with integer constant term. Note that $(p)$ is a maximal ideal of $S$ with $S/(p)\cong\mathbb{F}_p$. Let $R$ be the localization $S_{(p)}$. It is clear that $R$ is a local domain of characteristic $0$ with maximal ideal generated by $p$ and $R/(p)\cong \mathbb{F}_p$ is finite.
I claim, however, that $R$ is not Noetherian. Indeed, consider the chain of ideals $(x)\subset(x/p)\subset(x/p^2)\subset\dots$. These inclusions are all strict since $1/p\not\in R$ so $x/p^{n+1}$ is not a multiple of $x/p^n$ for any $n$.
Here is a very general principle along these lines. Let $R$ be any non-Artinian ring. Then I claim $R$ has an elementary extension (over the first-order language of rings) which is not Noetherian. Indeed, add a sequence of constant symbols $x_0,x_1,x_2,\dots$ to the language and axioms saying $x_n$ is not in the ideal generated by $x_0,\dots,x_{n-1}$ for each $n$. These axioms are finitely satisfiable in $R$, since $R$ is not Artinian so there are arbitrarily long finite chains of ideals in $R$. So by compactness, there is an elementary extension of $R$ that satisfies all of them, and thus is not Noetherian.
In particular, applying this with (say) $R=\mathbb{Z}_{(p)}$ this gives an elementary extension of $\mathbb{Z}_{(p)}$ that is not Noetherian. The properties of being local, characteristic $0$, $p$ being in the maximal ideal, a domain, and $R/(p)$ having $p$ elements are all first-order, and thus are still true of this elementary extension.
Best Answer
If you don't get how Zev's hint goes here are three more hints to apply Nakayama's Lemma:
Since $R$ is Noetherian it follows that $I$ is finitely generated as an $R$ - module.
Since $R$ is a local ring, $\operatorname{Jac}( R) = \mathfrak{m}$, the maximal ideal of $R$.
Every ideal is contained in some maximal ideal. The containment could be proper or we can have equality.