Rotate Δ δBPC 60 degrees counterclockwise around point B to get BAP', then BP'=BP=4, P'A=PC=5, connect PP'

∫∠PBP' = 60 degrees

∴δbpp' is equilateral,

∴PP'=PB=4

∫ 3 2+4 2 = 5 2, that is, PA 2+PP' 2 = P 'a 2.

∴∠APP'=90 degree

∠APB=90+60= 150。

(Not the size of ∠PAB)