In this answer I prove both points $1.$ and $2.$, let's start with point $1.$ The proof of point $1.$ only uses Hausdorff assumption.
Let $X$ be an extremally disconnected space and $p\in X$ be a non-isolated point of $X$. Let $(\mathscr{S}_\alpha)_{\alpha\in A}$ be a chain, where each $\mathscr{S}_\alpha$ is a family consisting of pairwise disjoint clopen subsets of $X$ that don't contain $p$. Their union $\bigcup_{\alpha\in A}\mathscr{S}_\alpha$ consists of pairwise disjoint clopen subsets of $X$ that don't contain $p$ as well. From Zorn's lemma we can take maximal such family $\mathscr{S}$. If $x\notin \overline{\bigcup\mathscr{S}}$ and $x\neq p$, then there exists a clopen set $U\ni x$ disjoint from $\bigcup\mathscr{S}$ and such that $p\notin U$ (Hausdorff extremally disconnected space is totally separated), which is impossible from maximality of $\mathscr{S}$. It follows that $X\setminus\{p\}\subseteq \overline{\bigcup\mathscr{S}}$ and so $\overline{\bigcup\mathscr{S}} = X$ since $X\setminus\{p\}$ is not closed.
Let $\mathfrak{U} = \{\mathscr{T}\subseteq\mathscr{S} : p\in\overline{\bigcup\mathscr{T}} \}$. It follows from the extremally disconnected property of $X$ that $\mathfrak{U}$ is a filter on $\mathscr{S}$. If $\mathscr{T}\notin \mathfrak{U}$, then since $\overline{\bigcup\mathscr{T}}\cup\overline{\bigcup\left(\mathscr{S}\setminus \mathscr{T}\right)} = \overline{\bigcup\mathscr{S}}\ni p$, we have $\mathscr{S}\setminus \mathscr{T}\in \mathfrak{U}$, and so $\mathfrak{U}$ is an ultrafilter. If $U\in\mathscr{S}$ then $p\notin U$ and so $\{U\}\notin\mathfrak{U}$, from which it follows that $\mathfrak{U}$ is a free ultrafilter on $\mathscr{S}$.
Give $\mathscr{S}$ discrete topology, so that $\mathfrak{U}\in\beta \mathscr{S}\setminus\mathscr{S}$. If $|X|$ is smaller than the first measurable cardinal, then $|\mathscr{S}|\leq 2^{|X|}$ is as also, and so $\mathscr{S}$ is realcompact. It follows that there exists bounded continuous function $\varphi:\mathscr{S}\to (0, 1]$ with $\varphi^\beta(\mathfrak{U}) = 0$, where $\varphi^\beta:\beta\mathscr{S}\to [0, 1]$ is the continuous extension of $\varphi$ to $\beta\mathscr{S}$ (see $8.8$ of Rings of continuous functions by Gillman and Jerison). A neighbourhood of $\mathfrak{U}$ in $\mathscr{S}\cup \{\mathfrak{U}\}$ is of the form $\mathscr{T}\cup \{\mathfrak{U}\}$ for some $\mathscr{T}\in\mathfrak{U}$ (see $6E.6$ of Gillman and Jerison), so for $n\neq 0$ there exists $\mathscr{T}_n\in \mathfrak{U}$ with $\varphi(T) < \frac{1}{n}$ for all $T\in\mathscr{T}_n$.
Define $f:\bigcup\mathscr{S}\to\mathbb{R}$ by $f(s) = \varphi(S)$ where $s\in S\in\mathscr{S}$, then $f$ is well-defined, bounded and continuous. Since open sets in an extremally disconnected space are $C^\ast$-embedded, we can extend $f$ continuously to $X$.
Since $|f(t)| < 1/n$ for $t\in \bigcup\mathscr{T}_n$ and $p\in\overline{\bigcup\mathscr{T}_n}$, it follows that $|f(p)|\leq 1/n$ for all $n$, and so $f(p) = 0$. But the $G_\delta$-set $\{x\in X : f(x) = 0\}$ does not intersect $\bigcup\mathscr{S}$, and so is not open. So $X$ is not a $P$-space.
Now let's prove point $2.$
Suppose that $X$ is a discrete space with $|X|$ larger or equal to the first measurable cardinal. Then $X$ is not realcompact, so its Hewitt realcompactification $\nu X$ is not equal to $X$, and so is not discrete. Since $X$ is a $P$-space, $\nu X$ is a $P$-space as well, and because $X$ is extremally disconnected, so is $\beta X$. Since $\nu X$ is a dense subspace of $\beta X$, $\nu X$ is extremally disconnected.
It follows that $\nu X$ is a non-discrete extremally disconnected $P$-space.
