The following is from the book Design of Analog CMOS Integrated Circuit, Page 180, Example 6.5
compute the transfer function of the common-gate stage.
$$\omega_{in} = \left[(C_{GS} + C_{SB}) (R_S \parallel R_{in}) \right]^{-1}$$
$$\Rightarrow \omega_{in} = \left[(C_{GS} + C_{SB}) (R_S \parallel \frac{1} {g_m+g_{mb}}) \right]^{-1}$$
$$\omega_{out} = \left[(C_{DG} + C_{DB}) R_D \right]^{-1}$$
The book consider input resistance for \$\omega_{in}\$.
$$R_{in} = \frac{1} {g_m+g_{mb}}$$
However, I think we should also include \$R_{out}\$ for \$\omega_{out}\$.
The following is the output resistance. Neglecting channel-length modulation, \$r_{out}\$ should be removed from the following hybrid \$\pi\$ model.
$$R_{out} = \frac{V_X} {I_X} = R_S$$
$$\omega_{out} = \left[(C_{DG} + C_{DB}) (R_D \parallel R_{out})\right]^{-1}$$
$$\Rightarrow \omega_{out} = \left[(C_{DG} + C_{DB}) (R_D \parallel R_{S})\right]^{-1}$$