I can do this problem. ∠BPC= 135 degrees, the key to this problem is the particularity of the data. The detailed explanation is as follows:

Rotate △CPB 90 degrees counterclockwise around point C to become △CQB. Connecting PQ, it is easy to get that △CQP is an isosceles right triangle. QP=4 root number 2, and because AQ=BP=2, AQ +PQ =AP, so ∠AQP=90 degrees, because ∠CQP=45 degrees, so ∠CQA= 135 degrees, and because △CQA and △CPB are congruent.

Thank you for your adoption.