The logic circuit shown above is equivalent to :

Truth table of the given gate:

$\begin{array}{lll} A & B & C \\ 0& 0& 0\\ 0& 1& 1\\ 1& 0& 0\\ 1& 1& 0\end{array}$

Truth table of option $\left(1\right)$

$\begin{array}{lll} A & B & C \\ 0& 0& 1\\ 0& 1& 1\\ 1& 0& 0\\ 1& 1& 1\end{array}$

Truth table of option $\left(2\right)$

$\begin{array}{lll} A & B & C \\ 0& 0& 1\\ 0& 1& 0\\ 1& 0& 1\\ 1& 1& 1\end{array}$

Truth table of option $\left(3\right)$

$\begin{array}{lll} A & B & C\\ 0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\\ 1& 1& 0\end{array}$

Truth table of option $\left(4\right)$

$\begin{array}{lll} A & B & C\\ 0& 0& 0\\ 0& 1& 1\\ 1& 0& 0\\ 1& 1& 0\end{array}$

Since option $\left(1\right)$ has same truth table, hence answer is option $\left(4\right)$ is correct.

Given Boolean expression can be written as

$\overline{A+\overline{B}}=C$

$\therefore C=\overline{A}·\overline{B}=\overline{A}·B$