The circuit given below is that of:

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(C) Let the inputs be $A$ and $B$. The first gate is an $OR$ gate with one input inverted,so its output $Y' = \bar{A} + B$. The next gate is an $AND$

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$NAND$ gate

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(C) Let the inputs be $A$ and $B$. The first gate is an $OR$ gate with one input inverted,so its output $Y' = \bar{A} + B$. The next gate is an $AND$ gate with inputs $Y'$,$A$,and $B$. Its output $Y'' = Y' \cdot A \cdot B = (\bar{A} + B) \cdot A \cdot B$. Using Boolean algebra: $Y'' = (\bar{A} \cdot A \cdot B) + (B \cdot A \cdot B) = 0 + A \cdot B = A \cdot B$. The final gate is a $NOT$ gate,so the final output $Y = \overline{Y''} = \overline{A \cdot B}$. This is the Boolean expression for a $NAND$ gate.

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