d
Let acceleration of wedge is \(a_{1}\) and acceleration of block w.r.t. wedge is \(a_{2}\)

\({N} \cos 60^{\circ}={Ma}_{1}=16 {a}_{1}\)

\(\Rightarrow {N}=32 {a}_{1}\)

F.B.D. of block w.r.t wedge

\(\perp\) to incline

\({N}=8 {g} \cos 30^{\circ}-8 {a}_{1} \sin 30^{\circ} \Rightarrow 32 {a}_{1}=4 \sqrt{3} {g}-4 {a}_{1}\)

\(\Rightarrow {a}_{1}=\frac{\sqrt{3}}{9} {g}\)

Along incline

\(8 g \sin 30^{\circ}+8 a_{1} \cos 30^{\circ}=m a_{2}=8 a_{2}\)

\(a_{2}=g \times \frac{1}{2}+\frac{\sqrt{3}}{9} g \cdot \frac{\sqrt{3}}{2}=\frac{2 g}{3}\)