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

$\mathrm{\Delta }p= p_f-p_i$

$=- \frac{E}{c}- \frac{E}{c}= -\frac{2E}{c}$

Thus , momentum transferred to the surface is

$\Delta p^{\prime }=|∆p|=\frac{2E}{c}$