Let $\kappa$ be a measurable cardinal with $\sigma$-complete ultrafilter $U$.  Let $X$ be the set of all finite sequences from $\kappa$. For $s\in X$, $i\in \kappa$, we write $(s,i)$ for the sequence you get by appending $i$ to $s$, similarly $(s,i,j)$, etc.  We call a subset $A \subseteq X$ closed if it has the following property: 

* Whenever $s$ in $X$, and almost all successors of $s$ are in $A$, then also $s$ is in $A$: 

More precisely: If the set $\{i \in \kappa: (s,i)\in A\}$is in $U$, then also $s\in A$. 

[EDITED:] In other words: A set $O$ is open if for all $s\in O$ also almost all successors of $s$ are in $O$. 
<strike>(Using the countable completeness of $U$, a neighborhood base of $s$ is given by the sets $O_{s,F}:=\{s\}\cup \{(s,i,j,\ldots, k): i,j,\ldots ,k\in F\}$, for $F\in U$. Using the fact that $U$ is non-principal one can show that these sets are clopen.)</strike>(DELETED, see Joseph's comment below.)

We check that $X$ is extremally disconnected: If $O$ is open, and $A$ is the closure of $O$, we claim that $A$ is open.  So let $s\in A$. 
If $s\in O$, then $s$ has a neighborhood in $A$, and we are done.   So assume that $s$ is not in $O$.  Then almost all successors of $s$ must also be  in $A$, otherwise $A\setminus \{s\}$ is closed. So $A$ is open. 

The fact that $X$ is a p-space follows from the countable closure of $U$. It is clear that $X$ has no isolated points.