Q: Help with making sure a correct use of a BJT I'm designing a PCB that receives 3 power inputs 3.3v, 5v and 12v.
I need simple way to switch all of them of in one go.
Here is a schematic of my PCB

My 12V line is for running a ws2815 led strip and may go up to 10A.
The 5V line is just for the level shifter.
The 3.3V is for running a NodeMCU ESP-12E board.
I'm using EVGA SuperNOVA 650 G1+ to drive everything  .
I can switch the outside power source but it should run a couple of this boards and I want to be able to Independently shut them off.
I need to run a lot of those and I'm already running 7A on the 12v line so I can't just keep drawing power form only the 12v line there is the matter of the cable and how much amp it can run.
I can use what ever components I want I just need to keep it small.    
Using help form an answer on a previous post of mine see here I came up with this switching solution:
 
Is this a correct use of BJT?
Do I need to add fly-back diode or a resistor somewhere?  
This is just a test circuit to see if it even works everything at default and the diode is there just a place holder. 

A: It's not ideal. The bases of the transistors will be pulled up to +11.3V by Q2 E-B junction so Q1 will see a reverse bias of 8.7V on the B-E junction, which violates the 5V absolute maximum value and is at the edge of where the B-E junction typically breaks down. 
You would be better to use NPN transistors rather than PNP with 3 base resistors. That will eliminate the problem mentioned above, and M12 will get more gate voltage when "on". It may not be adequate even with that but I can't analyze it because there is yet another problem- the LED currents are uncontrolled because you didn't add the required series resistors to set the currents. As it is, it will likely destroy the LEDs. 

A: I agree that you do not need the bipolar transistors, but you cannot simply tie the three MOSFET gates together.  The three gate pull-up (turn-off) resistors go to three different voltages.  Without some isolation, the pull up resistors form voltage dividers.  The 3.3 V source could turn on the 12 V FET.
Replace each 3906 with a 1N914 or 1N4148 diode.  Anodes to the individual gates, cathodes tied together to the switch.  Low Vgs might be an issue for the FET switching the 3.3 V source.  in that case ...
Is the unswitched 12 V source available all the time?  If yes, then the 12 V can be used to turn off all three FET gates.  Now you don't need the isolation diodes.

A: I think you could simplify it to this.  You'll have to find MOSFETs that will work at the voltages you had, especially some that will turn on/off with just 3.3V.

simulate this circuit – Schematic created using CircuitLab
