As we know, acceleration of the center of mass of cylinder rolling down an inclined plane

$a_{c}=\frac{g \sin \theta}{1+\frac{I}{M R^{2}}}$

In case of $P$ the mass is concentrated away from the axis, So $I_{P}>I_{Q}$

$\therefore \quad a_{P} < a_{Q} \Rightarrow v_{\mathrm{P}} < v_{\mathrm{Q}} \Rightarrow \omega_{\mathrm{P}} < \omega_{\mathrm{Q}}$