Simple Pendulum question from MHT CET 2020 (16 Oct, Shift 1)

Answer: (d)

1. For a simple pendulum $\omega^2 = \frac{g}{L}$.
2. At the extreme position all energy is potential: $PE = \frac{1}{2}m\omega^2A^2$.
3. $PE = \frac{1}{2}m\cdot\frac{g}{L}\cdot A^2 = \frac{mgA^2}{2L}$
4. So the answer is (d) $\frac{mgA^2}{2L}$.