Find the correct combination of A, B, C and D inputs which can...

The LED will glow if it is forward biased, which means the potential at the p-side must be higher than the potential at the n-side.
Let $Y_1$ be the output of the upper NAND gate (p-side)$and Y_2$ be the output of the lower NOR gate (n-side).
For the LED to glow, we need $Y_1 = 1$ and $Y_2 = 0$.
The upper circuit consists of two NOT gates (NOR gates with shorted inputs) followed by a NAND gate.
Output of first NOT gate is $\bar{A}$, and second is $\bar{B}$.
$Y_1 = \overline{\bar{A} \cdot \bar{B}} = \overline{\bar{A}} + \overline{\bar{B}} = A + B$ (using De Morgan's Law).
The lower circuit is a NOR gate with inputs $C$ and $D$.
$Y_2 = \overline{C + D}$.
Condition 1: $Y_1 = 1 \implies A + B = 1$. This means at least one of $A$ or $B$ must be 1.
Condition 2: $Y_2 = 0 \implies \overline{C + D} = 0 \implies C + D = 1$. This means at least one of $C$ or $D$ must be 1.
Checking the options:
(1) $A=1, B=1, C=0, D=1$: $Y_1 = 1+1 = 1$ and $Y_2 = \overline{0+1} = 0$. (Satisfied)
(2) $A=0, B=0, C=1, D=1$: $Y_1 = 0+0 = 0$. (Not satisfied)
(3) $A=1, B=0, C=0, D=0$: $Y_2 = \overline{0+0} = 1$. (Not satisfied)
(4) $A=0, B=1, C=0, D=0$: $Y_2 = \overline{0+0} = 1$. (Not satisfied)
Thus, the correct combination is 1101.