Equilateral Triangle Challenge

Rotate the triangle $60^\circ$ counterclockwise about point $B,$ so that $A\rightarrow A',$ $C\rightarrow C' = A,$ and $P \rightarrow P'.$ $BP = BP' = PP'$ because it forms an isosceles triangle (since $BP' = BP$ by symmetry) with a 60 degrees vertex angle (since the rotation was $60^\circ$ ), so it's equilateral.

Therefore, $AP' = CP$ , $P'P = BP,$ and $AP =AP$ , so they form a triangle with the three side lengths $CP,BP,AP$ which is given as a right triangle, so $AP'P = 90^\circ$ .

Thus, $\angle BPC = \angle BP'A = \angle BP'P + \angle PP'A = 60^\circ + 90^\circ = 150^\circ.$