Electronic – Small signal diagram for a PMOS

No, replace Vgs with Vsg. The models for a PMOS and NMOS transistor are the same, except for PMOS you use Vsg instead of Vgs, Vsd instead of Vds, and the absolute value of \$\lambda\$ and \$V_a\$.

For instance, in a velocity-limited model (\$I_d=V_{sat} C_{ox} W_g (V_{sg}-|V_t|) (1+|\lambda| V_{sd})\$), the transconductance would be \$V_{sat}C_{ox}W_g(1+|\lambda| V_{sg})\$. In a mobility limited model it would be \$ \mu C_{ox} \frac{W_g}{L_g} (V_{sg}-|V_t|) (1+|\lambda| V_{sd})\$.

These results makes sense, because in a PMOS transistor the current is proportional to Vsg (or negative Vgs). As you can see, in both these expressions, the transconductance is positively proportional to Vsg, and thus the current changes with Vsg.

The only real difference is in the mobility of the carriers, where electrons are faster than holes (around 2x) thus giving nMOS transistor a 2 times better performance for everything else equal.

About notation, it's mostly a practical thing, and the small signal model is an abstraction that you make to analyze a circuit. So you can use it the way you understand it better.

For instance, I usually use magnitudes (flipping notations drives me crazy) and then determine the direction of the current and voltage from the circuit. So the nMOS will generally have a current from drain to source, and the pMOS from source to drain, both with positive sign.

pMOS's \$I_{SD} \$ will be proportional to \$V_{SG} - |V_T| \$, where nMOS's \$I_{DS} \$ will be proportional to \$V_{GS} - |V_T| \$. Note that for the pMOS, you can flip SG and SD and still obtain the right values, as long as you use the absolute value of Vt.