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3$^{rd}$ NOR gate can replaced by Bubbled AND gate ==> those bubble will cancel the Bubbles of 1,2 gates.

∴ 1$^{st}$ is AND gate, 2$^{nd}$ is OR gate, 3$^{rd}$ is AND gate.

from 1$^{st}$ gate o/p is  = AB

from 2$^{nd}$ gate o/p is  = $\overline{A}$ + B

==> o/ of 3$^{rd}$ gate = AB . ( $\overline{A}$ + B ) = AB
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$Y=\overline{\overline{AB}+(\overline{\bar {A}+B}})$

$\implies Y=\overline{\overline{(AB)}}.(\overline{\overline{\bar A+B}})$

$\implies Y= AB.(\bar A+B)$

$\implies Y=AB.\bar A+AB.B$

$\because A.\bar A=0$

$\implies Y=AB$

Option $(C)$ is correct.
Answer:

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