Somewhere I am grasping the information that work done by static friction is zero as the point of contact where it is acting has zero velocity so it is at rest w.r.t ground
This is correct. $P=\vec F \cdot \vec v$ and since $\vec v = 0$ we get $P=0$ also.
Note that power is frame variant, meaning that it is different in different reference frames. If we transform to a different frame, indicated by prime marks, we can write $P'=\vec F \cdot v'$. If the primed frame is moving with respect to the unprimed frame at a velocity $\vec u$ then $\vec v' = \vec v - \vec u$ so $P'=\vec F \cdot (\vec v - \vec u) = P - \vec F \cdot \vec u$
why is the work done by the torque produced due to static friction zero?
This is due to the way that power transforms between reference frames, as described above. For rotational motion $P = \tau \omega$, where $\tau$ is the torque and $\omega$ is the angular velocity. But this formula applies only in the frame where the axis is at rest. In this problem the axis is moving at some velocity, $u$. So the power in this frame is $P' = P - \vec F \cdot \vec u = \tau \omega - F u = (F r) (u/r) - F u = 0 $
So when you correctly calculate the power from the torque in this reference frame you find that it is also zero.