Voltage Divider to protect P-FET gate on a Polarity Inversion Circuit
The Zener is better for two reasons:-


The Gate is clamped to a safe voltage no matter what the input voltage, whereas with a resistor divider Gate voltage continues to increase as input voltage increases.
The Zener doesn't reduce Gate voltage when below its Zener voltage, so the FET will still be well turned on at low input voltages. 


With resistors you need to balance the required Gate turn-on voltage against the voltage reduction required for protection. That limits the range of input voltages that the circuit can handle, and requires carefully chosen resistor values.

With a Zener, the resistor value just has to be high enough to not overheat the Zener or resistor, but low enough to provide adequate Zener bias current and to discharge the Gate rapidly if the supply voltage is suddenly reversed (so 10K is OK, but 10M might not be). 

The 1N4740A is rated for 25mA nominal, but should work down to a bit less than 1mA. At 30V in the resistor has to drop 20V, so its value could range from at least 20V/25mA = 800&ohm; to 20V/1mA = 20k. At 10V in the FET will get almost the whole 10V so it will still be fully turned on.

With a resistor divider, to get 10V on the Gate one resistor has to drop 20V and the other 10V, thus dividing the input voltage by 3. The lower resistor then has to be twice the value of the upper one, ie. if R1 is 10k then (the resistor in place of) D1 must be 5k. 

However at 10V the resistors will still divide the voltage by 3 so the FET will only get 3.3V - not enough to turn it on properly. This could be bad news if the power supply 'browns out' or the load draws a high surge current that momentarily drops the input voltage, as the partially turned on FET could dissipate high power and blow up.