In case of pure rolling on inclined plane,

$$\begin{aligned} & a=\frac{g \sin \theta}{1+I / m R^2} \\ & I_{\text {solid }} < I_{\text {hollow }} \\ \therefore \quad & a_{\text {solid }}>a_{\text {hollow }} \end{aligned}$$

$\therefore$ solid cylinder will reach the bottom first.
Further, in case of pure rolling on stationary ground, work done by friction is zero. Therefore, mechanical energy of both the cylinders will remain constant.

$$\begin{aligned} & \therefore(\mathrm{KE})_{\text {Hollow }}=(\mathrm{KE})_{\text {solid }}=\text { decrease in } \mathrm{PE}=m g h \\ & \therefore \text { correct option is }(\mathrm{d}) \end{aligned}$$