For a pulley and block system on a smooth double inclined plane as shown below


Force equation for both the blocks,
\(\Rightarrow \quad m g \cos 30^{\circ}-T=m a\) ...(i)
\(\Rightarrow \quad T-m g \cos 60^{\circ}=m a\) ...(ii)
From above equation we get,
\(a=\frac{(\sqrt{3}-1)}{4} g\)
\(\because\) Magnitude of the acceleration of centre of mass,
\(\mathbf{a}_{C M}=\left|\frac{m \mathbf{a}_1+m \mathbf{a}_2}{m+m}\right|\)
Here, \(\mathbf{a}_{C M}=\left|\frac{\left(\frac{\sqrt{3}-1}{4}\right) g \hat{\mathbf{i}}+\left(\frac{\sqrt{3}-1}{4}\right) g \hat{\mathbf{j}}}{2}\right|\)
\(\begin{aligned}
& =\frac{g}{2} \sqrt{\left(\frac{\sqrt{3}-1}{4}\right)^2+\left(\frac{\sqrt{3}-1}{4}\right)^2} \\
& =\frac{g}{2} \times \sqrt{2} \times \frac{\sqrt{3}-1}{4} \\
& =\left(\frac{\sqrt{3}-1}{4 \sqrt{2}}\right) g
\end{aligned}\)
Hence, option (d) is correct.