Drop-out voltage at the input using MOSFET
It does. But the capacitors hold some charge for a while before losing it slowly. Without capacitor voltage falls immediately. This is normal.
This resistor limits the current being sunk to ground. Without it you would create a short circuit situation. 


IMO this circuit is build upside down. The resistor should be on the drain side (to be a pull up resistor. Value between 10K to 50K at 10V) with all the capacitors and stuffs, and the source connected to GND directly. While the current setup works in practice, it's not intended to work like that.