Small-signal voltage gain of P-type common-gate circuit
Also, are you sure that \$K_P\$ is \$0.001\textrm{mA/V}^2\$? Because something is not quite right here. 

If for example, I use \$K_P = 1\textrm{mA}/V^2\$ instead. I've got \$I_{DQ} = 0.823\textrm{mA}\$ and 

\$g_m=2\sqrt{K_pI_{DQ}} \approx 1.81\textrm{mS}\$   

And the voltage gain:
$$A_V = g_m \times R_D||R_L = 1.81\textrm{mS} \times \frac{4\textrm{k}\Omega}{3} = 2.41\: V/V$$

And the input impedance \$R_{in} = \frac{1}{g_m }|| R_S = 485\Omega \$