The product of the reaction between ethyl benzene and N-bromosuccinamide is

N-Bromosuccinimide (NBS) in the presence of light or a peroxide is a reagent for selective radical bromination at an allylic or benzylic position. It maintains a very low concentration of molecular $$Br_2$$ so that substitution (not addition) occurs.

Ethyl benzene is $$C_6H_5CH_2CH_3$$. The carbon that is directly attached to the ring (the $$CH_2$$ group) is a benzylic carbon. Benzylic radicals are strongly resonance-stabilised, so the hydrogen atoms on this carbon are the ones that react fastest with NBS.

Step-wise outline:
1. Abstraction of a benzylic H gives the benzylic radical $$C_6H_5\dot{C}HCH_3$$.
2. The radical is trapped by $$Br_2$$ (generated in situ from NBS) to give the benzylic bromide and a new $$Br\cdot$$ radical, propagating the chain.

The overall transformation is therefore

$$C_6H_5CH_2CH_3 \;\xrightarrow[\;h\nu\;]{\;NBS\;} C_6H_5CH(Br)CH_3 + \text{succinimide}$$

The product $$C_6H_5CH(Br)CH_3$$ is named 1-bromo-1-phenylethane (also called α-bromoethylbenzene).

Hence the correct choice is
Option D which is: 1-bromo-1-phenylethane.