If someone doesn't get back to you before tonight, I'll do this when I'm online again.
Call this state . Now we know that
where K(x) is the kinetic energy operator.
Well, we know that , which is an even function, and we know that V(x) is given to us as an even function, so the only possible kind of eigenfunction we can have is an even function.
So the eigenstate is an even function. (We may easily extend this argument to an arbitrary function that satisfies the Schrodinger equation with an even V(x).)
On to the commutator:
But we know that , so
(the last term since E is simply a constant.)
and, again, , so
Since for an arbitrary function , thus [H, P] = 0.