d
When the string is cut, the rod will rotate about \(P.\) Let \(\alpha \) be initial angular acceleration of the rod. Then 

Torque, \(\tau  = I\alpha  = \frac{{M{L^2}}}{3}\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)\)

(Momentum of inertia of the rod about one end = \(\frac{{M{l^2}}}{3}\))

Also, \(\tau  = Mg\frac{L}{2}\)                                \( ...(ii)\)

Equating \((i)\) and \((ii)\), we get 

\(Mg\frac{L}{2} = \frac{{M{L^2}}}{3}\alpha \,\,or\,\,\alpha  = \frac{{3g}}{{2L}}\)