Angle in Equilateral Triangle and Square

In square $ABCD$, $AB = BC = CD = DA = a$. Since $BPC$ is equilateral, $BP = PC = BC = a$. $\angle PBC = 60^\circ \implies \angle ABP = 90^\circ - 60^\circ = 30^\circ$. In $\triangle ABP$, $AB = BP = a \implies \triangle ABP$ is isosceles. $\angle BAP = \angle BPA = \frac{180^\circ - 30^\circ}{2} = 75^\circ$. By symmetry, $\angle CPD = 75^\circ$. At point $P$, total angle $= 360^\circ = \angle BPA + \angle BPC + \angle CPD + \angle APD$. $75^\circ + 60^\circ + 75^\circ + \angle APD = 360^\circ \implies 210^\circ + \angle APD = 360^\circ \implies \angle APD = 150^\circ$.