special extremally disconnected spaces with only finite isolated points We Know that a cardinal $\kappa$ is measurable if there is a set $X$ with cardinal $\kappa$ and a {0,1}-measure $\mu: P(X) \rightarrow ${$0,1$} so that for all $x \in X$,  $\mu(x)=0$ and $\mu(X)=1$. also a cardinal which is not measurable, is called non measurable.
Also we Know  this is  unprovable to find a set with measurable cardinal in "ZFC".
In topology an extremally disconnected space is a topological space in which all open subsets has open closure.
Also we call a topological space to be a P-space if all it's $G_{\delta}$- sets are open.
There is a well-Known theorem that says every extremally disconnected P-space with non measurable cardinal is discrete.
From the aforesaid summaries a question could be posed that:

Question: If we suppose that a measurable cardinal exists, can we construct an extremally disconnected P-space with only a finite number of isolated points. 

 A: 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 $O$. 
(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.)(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. 
