The correct answer is option (E):
150°
Let the side length of the square ABCD be 's'. Since BPC is an equilateral triangle, BP = PC = BC = s. Also, angles ABC, BCD, CDA, and DAB are all 90 degrees.
Consider triangle ABP. AB = s and BP = s. Angle ABP = Angle ABC - Angle PBC = 90° - 60° = 30°. Since AB = BP, triangle ABP is an isosceles triangle. Therefore, angles BAP and BPA are equal. The sum of angles in a triangle is 180°, so angles BAP + angles BPA + angles ABP = 180°. Hence, 2 * angle BAP + 30° = 180°, which gives 2 * angle BAP = 150°, and angle BAP = 75°. Similarly, angle BPA = 75°.
Consider triangle CDP. CD = s and CP = s. Angle BCP = 60°. Then, Angle PCD = Angle BCD - Angle BCP = 90° - 60° = 30°. Since CD = CP, triangle CDP is also an isosceles triangle. Therefore, angles CDP and CPD are equal. The sum of angles in a triangle is 180°, so angles CDP + angles CPD + angles PCD = 180°. Hence, 2 * angle CDP + 30° = 180°, which gives 2 * angle CDP = 150°, and angle CDP = 75°. Similarly, angle CPD = 75°.
Now, we need to find angle APD. We can find it using the sum of angles around point P. Angle APD + angle BPA + angle BPC + angle CPD = 360°. Therefore, angle APD + 75° + 60° + 75° = 360°. Hence, angle APD + 210° = 360°. Thus, angle APD = 360° - 210° = 150°.