1. Use conservation of angular momentum

Since the two particles are attached symmetrically at diametrically opposite points, there is no external torque about the axis of rotation. Hence,

$L_i = L_f$

2. Initial moment of inertia

For a thin circular ring of mass $M$ and radius $r$ about its axis,

$I_i = Mr^2$

Initial angular momentum is

$L_i = I_i\omega = Mr^2\omega$

3. Final moment of inertia

Each attached particle has mass $m$ and is at distance $r$ from the axis.

So the two particles contribute

$2mr^2$

Therefore final moment of inertia is

$I_f = Mr^2 + 2mr^2 = (M+2m)r^2$

4. Apply conservation of angular momentum

Let the new angular speed be $\omega'$. Then

$Mr^2\omega = (M+2m)r^2\omega'$

Cancel $r^2$:

$M\omega = (M+2m)\omega'$

So,

$\omega' = \omega\frac{M}{M+2m}$

5. Match with options

This corresponds to:

$\boxed{\omega\frac{M}{M+2m}}$

So the correct option is C.