Why gmbs is not zero when B and S are tied together?
The value \$g_{mbs}\$ is a small-signal parameter. It is a linearization around a given operating point.

It answers the question how much the dependent variable would change if the independent variable is varied. Even though \$V_{BS}=0\$ we could change this voltage by a small amount \$v_{bs}\$ and consequently the drain current would change as well. The amount is given by \$v_{bs} g_{mbs}\$.

Likewise the \$g_m\$ of a transistor is not zero even if the gate is connected to a fixed voltage source.

Make a transient simulation of the following example and try to understand what's going on. I am sure the value of gmbs will suddenly make sense to you.



Update:
The reason why gmbs is not equal to zero lies in the fact that backgate voltage acts through the backgate-effect.

The drain current in saturation is given by the following equation
$$
I_D = \frac{K'}{2} \frac{W}{L} \left(V_{GS} - V_T(V_{BS})\right)^2
$$
where VT is a function of VBS!

Without going into the details of the derivation this finally results in
$$
g_{mbs} = g_m \frac{\gamma}{2\sqrt{2\Phi - V_{BS}}} = g_m \cdot \eta
$$

So, gmbs is proportional to gm!

Even if \$V_{BS}\$ is zero a small change will alter the threshold voltage an therefore the drain current changes as well.