Main power input section P-CH MOSFET connection confirmation
In your configuration, when the power is applied, MOSFET cannot be turned on immediately, because there's no voltage at the Source pin. So, current will flow first through the body diode until the voltage at the Source pin reaches the voltage level that turns the MOSFET on (You can see the forward-biased diode in your schematic). Once the MOSFET has turned on, it will short that diode and the current will flow through \$R_{DS-on}\$ in normal operation.

1) Of course. But if you don't connect the MOSFET that way, load current will always flow through body diode and the reverse polarity protection cannot be achieved.

2) Of course. But besides that, \$V_{GS}\$ is a deadly limitation for MOSFET. If supply voltage is 24V then the MOSFET will fail, because \$V_{GS-max}\$ is 12V for your MOSFET. A zener protection is required here.

Another thing which must be considered is power dissipation: \$P_M = I_D^2 \cdot R_{DS-on}\$. So the load current is important here. Your MOSFET has an on-resistance of \$R_{DS-on} = 15m\Omega\$ and it's in a SOT1220 package. And if the load current is 5A then \$P_M = 5^2 \cdot 0.015 = 0.3W\$. Thermal resistance, \$R_{th-ja}\$, of that package with proper copper-pad for drain is 67°C/W, so temperature rise will be around 20°C which causes no problems. But if the load current is 7A then temperature rise will be around 50°C -causing your MOSFET to get as hot as 75°C under room temperature!

Of course reverse polarity protection can be achieved with P-Ch. MOSFETs, but I personally recommend N-Ch MOSFETs because they've higher performance and lower price compared to P-Ch ones. Here's a configuration I'm using:







simulate this circuit &ndash; Schematic created using CircuitLab