In the figure shown, there is no friction between B and ground and $\mu =\frac{2}{3}$ between A and B.

The net work done on block A with respect to B is zero

The net work done on block B with respect to ground for a displacement ‘S’ is  $\frac{2MgS}{3}$

The net work done on block A with respect to ground for a displacement ‘S’ is  $\frac{MgS}{3}$

The work done by friction with respect to ground on A and B is equal and opposite in sign.