Thanks for your reply.
In Figure 1, the 12V is the Source pin of the P-channel MOSFET.
If the Gate is low, the P-channel MOSFET is switched on.
If the Drain is 12V, the Load is powered. But the Gate is still low and the Source is still 12V, keeping it switched on.
To slowly switch the load on, the Gate voltage has to be lowered slowly from 12V to a lower voltage.
Does that answer your question ?
I'm afraid not, no - the whole purpose of the solutions proposed in the app note are to ramp the MOSFETs on slowly so as to prevent high inrush currents.
If as you say, the gate is at 0 volts (because it's connected to the 0v rail), and you apply power to the circuit, i.e the source becomes 12V - then VGS is -12V, and the MOSFET is switched fully on, which then allows a large inrush current to flow immediately.
Hence, defeats the point entirely.