Solution: rotate △APB 90 around A to make AB and AC overlap and connect PP', then PA = P' A.

Then △APP' is an isosceles right triangle, then PP'=√PA? +P'A？ =√2

Autopythagorean Theorem: (√2)? +(√7)? =3? Is that PP? +PC？ =BP？ , then: △PP'C is also a right triangle, and ∠ p 'pc = 90, then: ∠ APC = ∠ app'+∠ p 'pc = 45+90 =135.

Otherwise, I can't work it out

I hope it inspires you.