The combination of gates shown below produces

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(D) The given circuit consists of two $NOT$ gates ($G_1$ and $G_2$) at the inputs $A$ and $B$,followed by a $NOR$ gate $(G_3)$ and another $NOT$ gate

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$NAND$ gate

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(D) The given circuit consists of two $NOT$ gates ($G_1$ and $G_2$) at the inputs $A$ and $B$,followed by a $NOR$ gate $(G_3)$ and another $NOT$ gate $(G_4)$.$1$. The outputs of $G_1$ and $G_2$ are $\bar{A}$ and $\bar{B}$ respectively.$2$. These are fed into the $NOR$ gate $G_3$,which produces the output: $Y' = \overline{\bar{A} + \bar{B}}$.$3$. Using De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.$4$. This output $Y'$ is then passed through the final $NOT$ gate $G_4$,resulting in $Y = \overline{Y'} = \overline{A \cdot B}$.$5$. The expression $\overline{A \cdot B}$ represents the output of a $NAND$ gate.Therefore,option $(d)$ is correct.

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