Mosfet gate driver have no direct connection to ground
In such cases, it is useful to redraw the circuit: as @Oldfart says in their comment, we have the following situation







simulate this circuit &ndash; Schematic created using CircuitLab

The source of the power MOSFET is connected to the ground via three of the four diodes \$D_1, D_2, D_3, D_4\$ packaged as a couple of double diodes: the resistors \$R_{21}\$ and \$R_{22}\$ bias this series connections of diodes with a approximately ten milliamperes, in order to let the source see few ohms of differential resistance (at least in the small signal regime) through ground and (perhaps this is the real reason of their presence) set a threshold of \$V_\mathrm{th} = 3V_\gamma+V_{GS_\mathrm{ON}}\$ for the gate driver signal voltage, where \$V_\gamma\$ is the threshold voltage of the diodes and \$V_{GS_\mathrm{ON}}\$ is the gate source turn-on voltage.

Edit
What happens if the source of the power MOSFET is connected directly to the ground, bypassing all the diodes and the capacitor \$C_{12}\$ with a short circuit? From my point of view, it is unlikely that this stage is used in some analog circuit, thus the only consequence I see is that, in this case, you simply have \$V_\mathrm{th} \simeq V_{GS_\mathrm{ON}}\$: this means that you obviously cannot set this threshold by changing the bias of \$D_1, D_2, D_3\$. Without knowing more details of the circuit, this is the only hypothesis I can do about how the circuit would work.