Level $1:$
- NAND gate Output are $: \overline{AB}$
- NOR gate Output are $: \overline{C+D}$
Level $2:$
- $1^{st}$ AND gate output are $: (\overline{AB})E$
- $2^{nd}$ AND gate output are $: EF$
- $3^{rd}$ AND gate output are $: F(\overline{C+D})$
Level $3:$
NOR gate Output are $:\overline{(\overline{AB})E+EF+F(\overline{C+D})}$
$\implies (AB + \overline{E}) \cdot (\overline{EF})\cdot (\overline{F} + C +D)\:\:\:[\because \text{De Morgan's law:} \:\:\overline{(A+B)} = \overline{A} \cdot \overline{B}\:\text{and}\: \overline{(AB)} = \overline{A} + \overline{B}\:]$
$\implies (AB + \overline{E}) \cdot (\overline{E}+ \overline{F})\cdot (\overline{F} + C +D)$
$\implies (AB\:\overline{E} + AB\:\overline{F} + \overline{E}\:\overline{E}+ \overline{E}\:\overline{F})\cdot (\overline{F} + C +D)$
$\implies (AB\:\overline{E} + AB\:\overline{F} +\overline{E} + \overline{E}\:\overline{F})\cdot (\overline{F} + C +D)$
$\implies [\overline{E}(AB +1 + \overline{F}) + AB\:\overline{F}] \cdot (\overline{F} + C +D)$
$\implies (\overline{E} + AB\:\overline{F}) \cdot (C + D + \overline{F} )\:\:\:\:[\because 1+X = 1]$
So, the correct answer is $(B).$