I see a few problems:

1. Biasing resistor of the 431. 330R brings a biasing current of 37 mA which is way too high. Normally, 1 mA is more than enough. Also it'll discharge the battery, let alone the power dissipation.
2. R4 is the pull-up of the comparator output, applies positive rail voltage to the MOSFET's gate by default but it's already shorted by the forward-biased diode, D1, so the R4 loses its "pull-up resistor" job.
3. Related to "2" above, if the comparator outputs 0V there's going to be a significant current flow through D1 as it'll short the battery rails (anode at +, cathode at – through comparator).
4. Look at the comparator arrangement again. If the battery voltage is higher than the threshold voltage which equals to Vbat = (33/8.2+1)*2.495 = 12.5V the comparator will output 0V. Now related to "3", the MOSFET will be off when the battery is charged enough, and since the output will be 0V (approx) the D1 may die due to overcurrent or even you may lose the battery completely.

So your circuit will not work as an over-discharge protection. It will overdischarge (or maybe even destroy) the battery. Even if you make the arrangement correctly, the unnecessary discharge due to biasing is still there. Also, if you do have to use these components then consider hysteresis with the comparator.

Finally, I'd recommend circuits and ICs specialised for this purpose, such as DW-01.

I see a few problems:

1. Biasing resistor of the 431. When the battery is charged enough i.e. have higher than 12.5 VDC, 330R brings a biasing current of more than 30 mA which is way too high already. Normally, 1 mA is more than enough. Also it'll discharge the battery, let alone the power dissipation.
2. R4 is the pull-up of the comparator output, and applies positive rail voltage to the MOSFET's gate by default but it's already shorted by the forward-biased diode, D1, so the R4 loses its "pull-up resistor" job.
3. Related to "2" above, if the comparator outputs 0V there's going to be a significant current flow through D1 as it'll short the battery rails (anode at +, cathode at – through comparator).
4. Look at the comparator arrangement again. If the battery voltage is higher than the threshold voltage which equals to \$\mathrm{V_{bat} = (33/8.2+1)\cdot 2.495 = 12.5V}\$, the comparator will output 0V i.e. pull the OUTPUT pin low and short to the negative rail. Now related to "3", the MOSFET will be off when the battery is charged enough, and since the output will be 0V (approx) the D1 may die due to overcurrent and fail-short which may possibly result in destruction of the battery.

So your circuit will not work as an over-discharge protection. It will overdischarge (or maybe even destroy) the battery. Even if you make the arrangement correctly, the unnecessary discharge due to biasing is still there. Also, if you do have to use these components then consider hysteresis with the comparator.

Finally, I'd recommend circuits and ICs specialised for this purpose, such as DW-01.