In the figure potential difference between A and B is:

8 V

The correct answer is Option (4) → 8 V

Battery Voltage = 24V

The Diode is in forward bias, so it will only behave as an wire.

$R_{eq}=(8+4)kΩ$

$=12kΩ$

and,

$I=\frac{V}{R}=\frac{24}{12kΩ}$  [Ohm's law]

$=2×10^{-3}Ω$

Voltage after $R_1$, $V=24-(8kΩ×2×10^{-3})V$

$=8V$

∴ Voltage at A, $V_A=8V$

and, 

Voltage at B, $V_B=8-\left(8×10^{-3}×\frac{8}{8×10^{-3}}\right)$

$=0$

$∴V_A-V_B=8V$