Consider the refraction at the face AB. Snell's law,

$$\sin i/\sin r = \sin 60^\circ /\sin r = \sqrt 3 $$,

gives $$r = 30^\circ $$.

The geometry in BCQP gives

$$\angle BPQ = 30^\circ + 90^\circ = 120^\circ $$

$$\angle CQP = 360^\circ - (135^\circ + 60^\circ + 120^\circ ) = 45^\circ $$.

Thus, the angle of incidence at Q is $${i_1} = 45^\circ $$. The critical angle for prism to air refraction is given by $$\sin {i_c} = 1/\sqrt 3 $$. Since $$\sin {i_1} = 1/\sqrt 2 > 1\sqrt 3 $$, we get $${i_1} > {i_c}$$ i.e., the angle of incidence is greater than the critical angle. Thus, the ray undergoes total internal reflection at Q. The laws of reflection gives $${r_1} = {i_1} = 45^\circ $$. In triangle QRD, $$\angle QRD = 60^\circ $$ and hence the angle of incidence at R is $${i_2} = 30^\circ $$.

Applying Snell's law at face AD, we get

$$\sqrt 3 \times \sin 30^\circ = 1 \times \sin e$$

or, $$\sqrt 3 \times {1 \over 2} = \sin e$$

$$\sin e = {{\sqrt 3 } \over 2}$$

or, $$e = {\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right) = 60^\circ $$

From figure,

The angle between the incident ray and the emergent ray is 90$$^\circ$$.

Hence, option (c) is correct and option (d) is incorrect.

Note : Angle between incident and emergent rays is the same as the angle between the two faces = 90$$^\circ$$.