Yes, this follows by properties of discrete valuations. Let $v_p$ and $\newcommand{\P}{\mathfrak P} v_\P$ be the (normalized) discrete valuations associated to the local rings $\mathbb{Z}_{(p)}$ and $\newcommand{\O}{\mathcal O} (\O_K)_\P$, respectively, where $\P = P_1$. (By normalized, I mean that $v_\P(a) = 1$ for all $a \in \P \setminus \P^2$, and similarly for $(p)$.) In general, letting $e = e_{\P|(p)}$ be the ramification index of $\P$ over $(p)$, we have $v_\P|_\mathbb{Q} = ev_p$ since $v_\P(p) = e$.
In your case $e = 1$, so $v_\P|_\mathbb{Q} = v_p$. Now
$$
p\mathbb{Z} = \{a \in \mathbb{Z} : v_p(a) \geq 1\}
$$
and in general
$$
p^n \mathbb{Z} = \{a \in \mathbb{Z} : v_p(a) \geq n\} \, .
$$
Similarly,
$
\P^n = \{a \in \O_K : v_\P(a) \geq n\} \, .
$
Since $v_p = v_\P$, then
$$
\P^n \cap \mathbb{Z} = \{a \in \mathbb{Z} : v_\P(a) \geq n\} = \{a \in \mathbb{Z} : v_p(a) \geq n\} = p^n \mathbb{Z} \, .
$$
I think this argument works for general Dedekind domains, too, since Dedekind domains are locally discrete valuation rings.
You can also see how this applies to your non-example, where $p \O_k = \P^2$. Then we have $v_\P|_\mathbb{Q} = 2 v_p$, so
$$
a \in \P^2 \cap \mathbb{Z} \iff 2 \leq v_\P(a) = 2v_p(a) \iff v_p(a) \geq 1 \iff a \in p \mathbb{Z} \, .
$$
Interestingly we also have $\P \cap \mathbb{Z} = p \mathbb{Z}$, since by the same logic as above, an element $a \in \P \cap \mathbb{Z}$ must satisfy $v_p(a) \geq 1/2$. But since $v_p$ takes values in $\mathbb{Z}$, this means that $v_p(a) \geq 1$. Thus in general, it looks like
$$
\P^n \cap \mathbb{Z} = p^{\lceil n/e \rceil} \mathbb{Z} \, .
$$
Addendum: The inclusion $\mathbb{Z} \hookrightarrow \O_K$ induces an inclusion of local rings $\mathbb{Z}_{(p)} \hookrightarrow (\O_K)_\P$. Let $\mathfrak{m} = p \mathbb{Z}_{(p)}$ and $\mathfrak{M} = \P (\O_K)_\P$ be their unique maximal ideals. Since $\mathbb{Z}$ and $\O_K$ are Dedekind domains, then $\mathbb{Z}_{(p)}$ and $(\O_K)_\P$ are discrete valuation rings. This means that $\newcommand{\m}{\mathfrak{m}} \newcommand{\M}{\mathfrak{M}} \m$ and $\M$ are principal, so $\M = (\pi)$ for some $\pi \in (\O_K)_\P$. Every element of $\mathbb{Z}_{(p)}$ can be written in the form $u p^a$ where $u \in (\mathbb{Z}_{(p)})^\times$ and $a \in \mathbb{Z}_{\geq 0}$ and every element of $(\O_K)_\P$ can be written as $v \pi^b$ with $v \in ((\O_K)_\P)^\times$ and $b \in \mathbb{Z}_{\geq 0}$. The discrete valuations associated to these are the exponents in these expressions, i.e., $v_\P(v \pi^b) = b$.
One definition of the ramification index $e = e_{\P|(p)}$ (when the extension of residue fields is separable, at least) is that it is the exponent such that $v_{\P}(p) = e$. Given $u p^a \in \mathbb{Z}_{(p)}$, then
$$
v_\P(u p^a) = a v_\P(p) = ae = e v_p(u p^a)
$$
which shows that $v_\P = e v_p$ on $\mathbb{Z}_{(p)}$, and since elements of $\mathbb{Q}$ can be written as quotients of elements of $\mathbb{Z}_{(p)}$, the same is true on $\mathbb{Q}$.
For more on discrete valuations, I recommend Chapter 1 of Serre's Corps Locaux and Chapter 7 of Rosen's Number Theory In Function Fields.