To determine the new angular velocity of the system, we use the principle of conservation of angular momentum. When no external torque acts on a system, its angular momentum remains constant. Let's denote the initial angular momentum and the final angular momentum, respectively, as $$L_{\text{initial}}$$ and $$L_{\text{final}}$$.

The initial angular momentum of the system is given by:

$$ L_{\text{initial}} = I_{\text{initial}} \cdot \omega $$

Here, $$I_{\text{initial}}$$ is the moment of inertia of the first disc. For a thin circular disc, the moment of inertia about its center is:

$$ I_{\text{initial}} = \frac{1}{2} M R^2 $$

Thus,

$$ L_{\text{initial}} = \left( \frac{1}{2} M R^2 \right) \omega $$

When the second disc is placed gently on the first disc, the two discs rotate together with a common angular velocity $$\omega'$$. The moment of inertia of the second disc is:

$$ I_{\text{second}} = \frac{1}{2} \left( \frac{M}{2} \right) R^2 = \frac{1}{4} M R^2 $$

The combined moment of inertia of the system after placing the second disc is:

$$ I_{\text{final}} = I_{\text{initial}} + I_{\text{second}} = \frac{1}{2} M R^2 + \frac{1}{4} M R^2 = \frac{3}{4} M R^2 $$

Thus, the final angular momentum of the system is:

$$ L_{\text{final}} = I_{\text{final}} \cdot \omega' = \left( \frac{3}{4} M R^2 \right) \omega' $$

By the conservation of angular momentum:

$$ L_{\text{initial}} = L_{\text{final}} $$

This simplifies to:

$$ \left( \frac{1}{2} M R^2 \right) \omega = \left( \frac{3}{4} M R^2 \right) \omega' $$

Solving for $$\omega'$$:

$$ \omega' = \frac{\left( \frac{1}{2} M R^2 \right) \omega}{\left( \frac{3}{4} M R^2 \right)} = \frac{\omega}{\frac{3}{2}} = \frac{2}{3} \omega $$

So, the new angular velocity of the system is:

$$\omega' = \frac{2}{3} \omega$$

Thus, the correct answer is:

Option D: $$\frac{2}{3} \omega$$