When is a P-space the set of P-points of a compact Hausdorff space? Let $X$ be a Hausdorff space. A point $x\in X$ is a P-point if every countable intersection of open neighbourhoods of $x$ is a neighbourhood of $x$, and $X$ is a
P-space if every point is P-point, equivalently if every countable intersection of open sets is open.
The set of P-points in a compact Hausdorff space is thus a completely regular P-space. My question is: is every completely regular P-space equal to the set of P-points of some compact Hausdorff space?
If not, is there a characterisation of those completely regular P-spaces which are equal to the set of P-points of some compact Hausdorff space?
Example: a countable discrete set $N$ is a P-space and is equal to the set of P-points of its one-point compactification $N\cup\{\infty\}$.
Sub-question: is every completely regular P-space $X$ equal to the set of P-points of its Stone-Cech compactification $\beta X$?
 A: The sub-question might not be true: 
Let $X$ be discrete, $\beta X$ be represented by the set of all ultrafilters on $X$. Then it is easy to see that an ultrafilter $\mathfrak{F}$ is a P-point in $\beta X$, iff $\mathfrak{F}$ has the countable intersection property.
Hence, if $X$ is of measurable cardinality, then $X$ does not equal the set of P-points of $\beta X$.
However, this does not provide a counter-example to the original question (consider the one-point compactification of the discrete space).  
Vice versa, if there does not exist a measurable cardinal, then the sub-question holds at least for discrete spaces.
A: Here is a complete answer for the sub question:
Let $X$ be completely regular.

*

*If $X$ is realcompact and $x_0$ is a P-point of $\beta X$, then $x_0 \in X$.

Proof:
Assume $x_0 \notin X$. Then there is a continuous
$h: \beta X \rightarrow [0,1]$ such that $h(x_0) = 0$ and $h(x) > 0$ for any $x \in X$
(Engelking, General Topology, Theorem 3.11.10).
Since $x_0$ is P-point, there is an open $U$ such that
$x_0 \in U \subseteq \bigcap_{n \in \textbf{N}} h^{-1} ([0, \frac{1}{n}[)$.
Hence there is an $x \in U \cap X \Rightarrow h(x) = 0$. Contradiction!


*Let $X$ be a P-space. Then:
$X$ is realcompact $\Leftrightarrow$ $X = \{x \in \beta X: x$  P-point in $ \beta X\}$
Proof:
"$\Rightarrow$":  "$\subseteq$":  Each P-point of $X$ is P-point in $\beta X$, since $X$ is dense in $\beta X$.
"$\supseteq$":  1.  
"$\Leftarrow$":
By Engelking, General Topology, Theorem 3.11.11 we have to show that an ultrafilter $\mathfrak{F}_0$ in
the zero-sets of $X$, which has the countable intersection property, has non-empty intersection.
Since in a P-space each zero-set is clopen, hence the complement is zero-set again, and each countable intersection of zero-sets is a zero-set,
it is easy to see that in fact each countable intersection of $\mathfrak{F}_0$ is an element of $\mathfrak{F}_0$. 
Now consider the standard construction of the Cech-Stone compactification
$\beta X = \{\mathfrak{F}: \mathfrak{F}$ zero-set ultrafilter in $X$}.
Since complements of zero-sets are zero-sets,
$\{ [A] : A$ is zero-set in $X \}$ is a (clopen) base of $\beta X$, where
$[A] := \{ \mathfrak{F} \in \beta X: A \in \mathfrak{F} \}$.   
$\mathfrak{F}_0$ is P-point in $\beta X$:
Let $(U_n)_{n \in \textbf{N}}$ be a sequence of neighbourhoods of $\mathfrak{F}_0$.
Then for each $n$ pick a zero-set $A_n$ of $X$ such that
$\mathfrak{F}_0 \in [A_n] \subseteq U_n$.
As stated above, $A := \bigcap_{n \in \textbf{N}} A_n \in \mathfrak{F}_0$.
Hence $\mathfrak{F}_0 \in [A] \subseteq  \bigcap_{n \in \textbf{N}} [A_n] \subseteq \bigcap_{n \in \textbf{N}} U_n$
and $[A]$ is open. 
By the prerequisite, $\mathfrak{F}_0 \in X$, hence  $\bigcap \mathfrak{F}_0 = \{x\}$ for some $x \in X$.
Notes

*

*Of course, this also gives an affirmative answer to the original question in case $X$ is realcompact.


*But there are non-realcompact P-spaces, for instance:  
a) any discrete space of measurable cardinality 
b) Example 9L in the book of
Gillman, Jerison,
which is $\{\alpha < \omega_2:$  cf $(\alpha) \neq \omega\}$.
Hence, this provides a ZFC counter-example to the sub question.
By the above, for these spaces $\beta X$ does not provide the required compactification.
But for a), the one-point compactification of the discrete space does the job.
However, in b) $\omega_2$ is a new P-point in the "obvious" compactification
$\omega_2 + 1$. Hence, this might be a candidate for a counter-example also of the original question.
