Why do we need NPN to drive PMOSFET in below circuit ? Can it be directly driven using uC.
Below both circuit is working but first is most popular.
If second circuit is working what is use of BC547 and R2 ?

 

... Just a guess, but the circuit with the NPN transistor used to pulse the mosfet gate may have the capability of draining the mosfet gate charge more rapidly than the direct version. Overheating can be a problem if the mosfet turns on and off too slowly, operating in a way that may not be optimal.
A breadboard test circuit might confirm this idea.

 

And the circuit with the transistor is used when the V1 supply is above the micro supply.
We tend to see quite a few posts here where folk try to drive the FET directly with higher V1 volts and that will not work.
The FET will be on all the time and the port will probably be popped.

 

In the circuit you show, any MCU at 3.3V can drive that mosfet directly. That is done all the time. Be sure to consider start up conditions when the MCU pins may not yet be defined.

 

And the circuit with the transistor is used when the V1 supply is above the micro supply.
We tend to see quite a few posts here where folk try to drive the FET directly with higher V1 volts and that will not work.
The FET will be on all the time and the port will probably be popped.

This is the real reason for using an NPN to drive the MOSFET.

 

Did you calculate the very slow gate risetime caused by the 47k resistor charging the high gate-source capacitance and the faster but still slow falltime caused by the fairly low current in the NPN transistor?

 

The P-MOSFET gate voltage must be essentially equal to the source voltage to fully turn off the MOSFET.
Thus, as noted, if the gate drive does not go as high as the source voltage, the MOSFET may not turn off.
In that case the NPN is used so that the gate voltage will always go to the source voltage (through R1) when the transistor is off.