Ultrafilters on naturals determine p-adic numbers

The set $\beta(\mathbb{N})$ has a natural topology that makes it the Stone-Cech compactification of $\mathbb{N}$, and your map $\Phi$ is just the unique continuous extension of the inclusion map $\mathbb{N}\to\hat{\mathbb{Z}}$. In particular, since $\Phi$ is continuous, its image is closed in $\hat{\mathbb{Z}}$. Since $\mathbb{N}$ is dense in $\hat{\mathbb{Z}}$, this implies $\Phi$ is surjective.

Concretely, given $a\in\hat{\mathbb{Z}}$ and nonzero $n\in\mathbb{Z}$, let $X_{a,n}$ be the set of integers that have the same mod $n$ residue as $a$. For fixed $a$, these sets $X_{a,n}$ generate a proper filter (since $X_{a,n}\cap X_{a,m}=X_{a,\operatorname{lcm}(m,n)}$) which can then be extended to an ultrafilter, and this ultrafilter maps to $a$ under $\Phi$.

Suppose that $a=\langle a_n:n\in\Bbb Z^+\rangle\in\prod_{n\in\Bbb Z^+}\Bbb Z/p^n\Bbb Z$ is in $\Bbb Z_p$, i.e., is such that $a_m\equiv a_n\pmod{p^m}$ whenever $m\le n$. For $n\in\Bbb Z^+$ let

$$A_n=\left\{k\in\Bbb Z^+:k\equiv a_n\pmod{p^n}\right\}\,;$$

if $m\le n$, then $A_n\subseteq A_m$, so there is a $\mathscr{U}\in\beta\Bbb N$ such that $A_n\in\mathscr{U}$ for each $n\in\Bbb Z^+$, and clearly $\big(\Phi(\mathscr{U})\big)(p)=a$. Thus, for each prime $p$ and each $p$-adic integer there is an ultrafilter that picks out that $p$-adic integer. (I don’t, however, know whether $\Phi$ is surjective.)