If point A is AC⊥ plane α and point C is on plane α and connects BC, then ∠ABC is the angle formed by straight line AB and plane α.

Now back to your question:

The point B of the straight line BP is outside the plane PEC, and the point P is on the plane PEC, so now it is necessary to cross the point B as the vertical line of the plane PEC.

Because it has been proved in the problem (1) that BC⊥ plane PEC, ∠BPC is the angle formed by straight line BP and plane PEC.

That is, ∠ BPC = 45 ①, and because the BC⊥ plane PEC and CP are on the BC⊥CP② plane PEC,

From ① ②, we can know that △BCP is an isosceles right triangle, and then it is easy to do it.