△BPC rotates 90 clockwise around point C, so that BC edge coincides with AC, and point P rotates to point P'

Then: △CPP' is an isosceles right triangle.

∴PP'^2=PC^2+P'C^2=2^2+2^2=8

P'A^2=PB^2= 1

PA^2=3^2=9

∴PP'^2+P'A^2=PA^2

∴∠AP'P=90

∴∠bpc=∠ap'c=∠ap'p+∠pp'c=90+45 = 135