In the pyramid $P$-$ABCD$, $AB \parallel CD$ and $\angle BAP = \angle CDP = 90 ^ { \circ }$.
(1) Prove that plane $PAB \perp$ plane $PAD$;
(2) If $PA = PD = AB = DC$ and $\angle APD = 90 ^ { \circ }$, find the cosine of the dihedral angle along edge $PD$ between plane $PAD$ and plane $PCD$.