Two small balls A and B , each with mass m , are rigidly attached...

Let the new centre of mass of the system after the collision be $C'$.
After the stationary particle $P$ sticks to ball $A$, the mass at $A$ becomes $m + m = 2m$, while the mass at $B$ remains $m$.
The distance of the new centre of mass $C'$ from $A$ is:

$x = \dfrac{m \times L + 2m \times 0}{3m} = \dfrac{L}{3}$

The distance of $C'$ from $B$ is $L - \dfrac{L}{3} = \dfrac{2L}{3}$.

The moment of inertia of the system about $C'$ after the collision is:

$I = 2m\left(\dfrac{L}{3}\right)^2 + m\left(\dfrac{2L}{3}\right)^2$

$I = \dfrac{2mL^2}{9} + \dfrac{4mL^2}{9} = \dfrac{2mL^2}{3}$

We apply the principle of conservation of angular momentum about $C'$ just before and after the collision. Before the collision, the angular momentum of the system about $C'$ is the sum of the angular momenta of balls $A$ and $B$ (since particle $P$ is stationary). Taking the direction of rotation into account:

$L_i = m v_0 \left(\dfrac{L}{3}\right) - m v_0 \left(\dfrac{2L}{3}\right) = -\dfrac{mv_0L}{3}$

The magnitude of the initial angular momentum is $L_i = \dfrac{mv_0L}{3}$.

After the collision, the system rotates about $C'$ with angular speed $\omega$. The final angular momentum is:

$L_f = I\omega = \left(\dfrac{2mL^2}{3}\right)\omega$

Equating the magnitudes of initial and final angular momentum ($L_i = L_f$):

$\dfrac{2mL^2}{3}\omega = \dfrac{mv_0L}{3}$

$\omega = \dfrac{v_0}{2L}$

Answer: $\dfrac{v_0}{2L}$