This answer uses a simplified method, which avoids the complications of having to deal with internal forces between the link and cylinders.

The motion of the two cylinders is identical, so they can be treated as one cylinder of mass $2M$. The link is then a point mass $m$ located at distance $r_0$ from the centre O. We can treat the link as a point mass because its orientation does not change throughout the motion. Like a point mass it has only translational KE, it has no rotational KE.

The moment of inertia of the double-cylinder about the point of contact with the ground P is

$$I=2(\frac12Mr^2+Mr^2)=3Mr^2$$ The moment of inertia of the double-cylinder and link about P is therefore $$I_P=3Mr^2+mr_1^2 \approx 3Mr^2+m(r-r_0)^2$$ where $r_1$ is the distance LP and for small oscillations $r_1 \approx r-r_0$.

The torque about the axis at P is $\tau\approx mgr_0\theta$ in the small angle approximation. This acts to decrease $\theta$ so the equation of motion is $$\tau=-I_p \ddot \theta$$ $$mgr_0\theta \approx -[3Mr^2+m(r-r_0)^2]\ddot\theta$$ $$\ddot\theta+\frac{mgr_0}{3Mr^2+m(r-r_0)^2}\theta \approx0$$ The natural frequency is $$f_n\approx 2\pi \sqrt{\frac{mgr_0}{3Mr^2+m(r-r_0)^2}}$$

The reaction forces at the hinge are more difficult. Draw FBDs for the cylinder and the link. The forces on the two cylinders are identical, so you need a diagram for only one. The hinge forces have horizontal and vertical components. Apply Newton's 2nd law to each body.

As a further simplification, imagine that the cylinders overlap and become one. You can do this because their motion is identical. Then the link has zero length - ie it is only a point mass. (The question does not tell you the length of the link, only its mass. So the length is irrelevant, and could be zero.) Then you have one cylinder with mass 2M and an off-centre mass m, rocking on the table. Now there is no need to consider the forces at the hinge.

edited May 6 by sammy gerbil