I am actually studying trees at the moment, and have been working on a slight variant of your problem, so I'll give it a shot. In what follows, denote by d(A,B) the distance (length of the path) from a vertex A to a vertex B.
Oddly, it seems the way do to this would be to prove 2) in order to prove 1). Proving 2) would go as follows: if A is any vertex on P(T), Let B be the vertex on P(T) such that a longest path in P(T) starting at A ends at B (so that A has eccentricity d(A,B)). We know that B is a leaf on P(T), for if it were not, we could just keep going and get a longer path from A. And since B is a leaf on P(T), T is gotten from P(T) by adjoing some number of new leaves to B (as well as possibly some other leaves on other vertices). Take any new leaf C connected to B, and we have that, in T, d(A,C) = d(A,B) + 1. We claim that d(A,C) is now the eccentricity of A in T. This is because, as we just argued, the distance from A to any other new leaf is simply the distance to that leaf's "branch" plus 1.
Thus, we've shown that the eccentricity of a vertex A inside T is equal to the eccentricity of A inside P(T), plus 1. So if the eccentricity of A inside P(T) is smaller than B's inside P(T), it must also be inside T as well.
We have to finally argue that none of the new leaves on T can qualify as centers for T. Let L be a leaf on T, and let B be the vertex (branch) that it is connected to. Then the eccentricity of L is equal to the eccentricity of B plus 1, so L immediately has greater eccentricity than another vertex in T. Thus L is not a center of T.
Thus we have proved 2).
Now, how to prove 1). I have not gone through this yet, but I'm pretty sure it can be proved using an induction argument. The trick is to realize that, starting with a tree T, and going to P(T), and then to P(P(T)), etc, you will eventually end up with a tree having exactly one or two vertices. The theorem is obviously true for such trees. And from part 2), we know that the center of T = center of P(T) = center of P(P(T)) = ... = center of the tree we finally end up with, which consists of 1 or 2 vertices.
Hope that was clear, let me know if any of it was not.