What is the role of the rightmost transitor in this circuit?
The pair of transistors form a "high side switch". The left (N-channel) transistor switches the gate voltage to the right (P-channel) transistor. The P-channel transistor requires a voltage that is negative with respect to the source (at +3.8V) in order to turn on. This allows the voltage to a grounded load to be switched, which is very important in some applications. 

In some cases, you might be able to leave out the N-channel part and reverse the logic (low = ON) but the gate voltage has to go up to very close to the 3.8V source to ensure the P-channel MOSFET is fully off, and 2V is far too low for that. 

The gate capacitor and series 50K/560K resistor is a bit more subtle- the load is presumed to have a lot of capacitance, so switching it very rapidly, as MOSFETs are wont to do, would cause a serious glitch in the 3.8V supply. Switching the MOSFET slowly allows time for the load capacitors to charge. This is very much situation-dependent, for example bringing the supply voltage up too slowly may cause some poorly designed circuits or chips to start up incorrectly. 

It's unclear why they would want to switch the MOSFET off so slowly (560+50K) vs. on (50K) time constant with the 10n + gate charge.