EDIT: The answer I gave was a bit too terse and left out some critical details which left a hole in the proof. Fortunately it can be salvaged with some caveats based on where the zeroes of the derivative lie which are described after the proof of the following lemma.
First let's prove a lemma that establishes an equality of sets,
$$P(\mathbb{Z}_p) = \bigcup_{x \in \mathbb{Z}_p} B_{\le |pP'(x)^2|}(P(x))$$
Here the notation $B_{\le r}(c) :=\{z \in \mathbb{Z}_p : |z-c|\le r\}$. Note that $r=0$ is allowed, except it's no longer an open ball, but the singleton, $B_{\le 0}(c) = \{c\}$.
We clearly see that because $P(x)$ is the center of every ball or singleton in the union, the image is contained within it:
$$P(\mathbb{Z}_p) \subseteq \bigcup_{x \in \mathbb{Z}_p} B_{\le |pP'(x)^2|}(P(x))$$
To show the other containment, take an arbitrary point of a ball, $y \in B_{\le |pP'(x)^2|}(P(x))$. This satisfies the inequality (when $P'(x) \ne 0$),
$$|P(x)-y|\le |pP'(x)^2| < |P'(x)|^2$$
If we pick $f(x)=P(x)-y$ we have that the general criteria for Hensel's lemma is satisfied for $|f(x)|<|f'(x)|^2$, and so $x_{n+1}=x_n-\frac{P(x_n)-y}{P'(x_n)}$ converges to some $z$ that makes $P(z)-y=0$, which proves $y=P(z) \in P(\mathbb{Z}_p)$. Just as a quick remark, the singleton case when $P'(x)=0$ forces the inequality $|P(x)-y|\le 0$ and so we have simply $P(x)=y$ and no Hensel lemma required.
Now that we have established this lemma, here are where the caveats will begin to apply. The singleton sets are precisely the points $\{P(x)\}$ where $P'(x)=0$. There are only finitely many of these problem points for any given polynomial, so some possibilities for these to not cause problems are when the only $x$ which gives $P'(x)=0$ lies outside $\mathbb{Z}_p$, possibly in $\mathbb{Q}_p$ or a finite extension (for instance $px^2+x$ or $x^3-x$), or the set $\{P(x)\}$ is contained in one of the open balls (I don't have an example of this last case off hand). Checking the Newton polygon of $P'$ is a simple way to confirming/creating any "nice" polynomial that meets the first criteria.
Supposing now that we have no problem singleton sets, the lemma shows $P(\mathbb{Z}_p)$ is a union of infinitely many open balls. Since $P$ is continuous and $\mathbb{Z}_p$ is compact, we know $P(\mathbb{Z}_p)$ is compact. These open balls form a cover that must contain some finite subcover by balls, of which there is some minimum radius. In other words, we can represent the image for some constant $p^{-n}$ and some centers $\{c_t\}_{t=1}^N$ as,
$$P(\mathbb{Z}_p) = \bigcup_{t=1}^N B_{\le p^{-n}}(c_t)$$
Specifically we can choose $c_t \in \{0,\dots,p^n-1\}$ and use this to compute $I_n=N$ because $P(x)=c_t \mod p^n$ for some p-adic integer $x$ which, by density of the integers in $\mathbb{Z}_p$ lets us pick a regular integer to do so. So our ratio is $\frac{I_n}{p^n} = \frac{N}{p^n}$.
We can further subdivide these balls evenly into the disjoint balls,
$$B_{\le p^{-n}}(c_t) = \bigcup_{a=0}^{p-1} B_{\le p^{-n-1}}(c_t+ap^n)$$
Each of these balls contains $p$ more integer points which gives us solutions to all of $P(x)=c_t+ap^n \mod p^{n+1}$ for each $a \in \{0,\dots,p-1\}$. Each of these $N$ balls when subdivided this way gets us $p$ times as many points, so we have $\frac{I_{n+1}}{p^{n+1}} = \frac{Np}{p^{n+1}} =\frac{N}{p^n}$. We can repeat the subdivision process and the numerator and denominator will continue to multiply by the equal powers of $p$, leaving us with,
$$\delta = \frac{N}{p^n}$$
Unfortunately the singleton sets where the derivative is 0 is a problem still, but maybe with a bit more work these can be resolved too.