To calculate the gain due to Vin,cm to X and Y, we use equivalent circuit in figure (b).

Now, $I_{d1} = gm_1 ( V_{in,cm} - V_p )$ and $I_{d2} = gm_2 ( V_{in,cm} - V_p )$

That is,

$(g_{m1}+g_{m2})(V_{in.CM}-V_p)R_{SS}=V_p$

and

$V_p = \frac{(g_{m1}+g_{m2})R_{SS}}{(g_{m1}+g_{m2})R_{SS}+1}V_{in.CM}$

We now obtain output voltages as

$V_x= -g_{m1}(V_{in.CM}-V_p)R_D$

$\hspace{0.5cm} = \frac{-g_{m1}}{(g_{m1}+g_{m2})R_{SS}+1}R_DV_{in.CM}$

and

$V_y= -g_{m2}(V_{in.CM}-V_p)R_D$

$\hspace{0.5cm} = \frac{-g_{m2}}{(g_{m1}+g_{m2})R_{SS}+1}R_DV_{in.CM}$

The differential component at the output is therefore given by

$V_x-V_y = -\frac{g_{m1}-g_{m2}}{(g_{m1}+g_{m2})R_{SS}+1}R_DV_{in.CM}$

In other words, the circuit converts input CM variations to a differential error by a factor equal to

$A_{CM-DM}=-\frac{\Delta g_m R_D}{(g_{m1}+g_{m2})R_{SS}+1}$

where $A_{CM-DM}$ denotes common-mode to differential mode conversion and $\Delta g_{m} = g_{m1} - g_{m2}$