I'm studying this circuit below which in fact uses a N-channel MOSFET in a power line, with source at the supply voltage and drain at the load.
Why the source is the input supply voltage and not the drain?
Because there is a parasitic diode inside the MOSFET and if you put it the other way it would not allow the input voltage to be much less than the output voltage (even on standby).
The MOSFET is equally happy allowing current to flow either way, even though it seems to be operating 'backward' when the 4.5V supply is present.
Edit: See the waveform during transition to standby:
If the MOSFET was the other way around, the standby power would backfeed other loads that may be connected to the 4.5V input supply.
This is the feature mentioned on the front page of the datasheet:
Input Disconnect Switch Isolates Input During Backup