Explanation

(b) Initial momentum of surface
${\ \ \ P}_{i} = \frac{E}{C}$
Where, $c\ $= velocity of light (constant).
Since, the surface is perfectly, reflecting, so the same momentum will be reflected completely.
Final momentum
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P_{f} = \frac{E}{c}$ (negative value)
$\therefore$Change in momentum
$\Delta p = \ p_{f} - p_{i}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - \ \frac{E}{c} - \ \frac{E}{c} = \ - \frac{2E}{c}$
Thus , momentum transferred to the surface is
$\Delta p^{'} = |\mathrm{\Delta}p| = \frac{2E}{c}$