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(B) The given circuit consists of an $OR$ gate where input $B$ is passed through a $NOT$ gate,and the output of the $OR$ gate is passed through anothe

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(B) The given circuit consists of an $OR$ gate where input $B$ is passed through a $NOT$ gate,and the output of the $OR$ gate is passed through another $NOT$ gate. Let the inputs be $A$ and $B$. The input to the $OR$ gate becomes $A$ and $\bar{B}$. The output of the $OR$ gate is $Y' = A + \bar{B}$. The final output $C$ is the inversion of $Y'$,so $C = \overline{A + \bar{B}}$. Using De Morgan's Law,$C = \bar{A} \cdot \overline{\bar{B}} = \bar{A} \cdot B$. This expression $\bar{A} \cdot B$ represents an $AND$ gate with an inverted input $A$.

The logic circuit shown above is equivalent to:

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