Consider the action of $P$ on $E$ defined by $g*x=gxg^{-1}$ for $g\in P$ and $x\in E$. This is well defined, since the property of being $p$-regular is clearly preserved by group isomorphisms. The set of fixed points under this action (i.e. the elements of $E$ left fixed by all elements of $P$) is $C\cap E$ by the very definition of the centralizer.
The proof is finished by the following
Lemma: Let $P$ be a $p$-group acting on a finite set $E$. Let $E^P$ be the set of fixed points under this action. Then $|E|\equiv|E^P|\text{ mod }p.$
Proof: We have to show that the order of $E\backslash E^P$ is a multiple of $p$. By definition, $E^P$ is the union of the one-point orbits of the action. So the order of $E\backslash E^P$ is the sum of the orbit orders $>1$. By the orbit-stabilizer-theorem, the order of an orbit divides the order of $P$. Since $P$ is a $p$-group, it follows that orbit orders $>1$ have to be multiples of $p$. So the order of $E\backslash E^P$ is the sum of multiples of $p$ and therefore itself a multiple of $p$.