Solution: rotate △BPC clockwise around point B to △BCP', or even P'P,

Obviously △BPP' is a right triangle, BP=BP'=2, ∠BDP'=45,

Therefore, in the right triangle BPP, from the Pythagorean theorem, PP'=2√2,

PA=2，CP'=AP= 1，

pp'^2=8,p'c^2= 1,pc^2=3^2=9，

PP'^2+P'C^2=PC^2

So △PP'C is a right triangle,

So ∠ PP 'c = 90,

So ∠BP'C=45+90= 135,

So ∠ APB = ∠ CP 'b = 135.