small signal analysis solve
It's easy to see that the small-signal drain voltage is given by

$$v_d = -i_d R_L||(R_1 + R_2)$$

so that's all there is to that.

If this were a true common-gate circuit, the small-signal gate-source voltage would be

$$v_{gs} = v_g - v_s = 0 - v_{sig} = - v_{sig}$$

but, in this circuit, the gate is not at signal common so we have

$$v_{gs} = v_g - v_s = v_d \frac{R_1}{R_1 + R_2} - v_{sig}$$

And, recalling that

$$i_d = g_mv_{gs}$$

you should have all you need to finish the problem.