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(C) The given circuit consists of a $NOR$ gate, a $NAND$ gate with its inputs tied together (acting as a $NOT$ gate), and a $NOT$ gate.
Let the input

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$NOR$

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(C) The given circuit consists of a $NOR$ gate, a $NAND$ gate with its inputs tied together (acting as a $NOT$ gate), and a $NOT$ gate.
Let the inputs be $A$ and $B$.
The output of the $NOR$ gate is $Y_1 = \overline{A+B}$.
The $NAND$ gate with tied inputs acts as a $NOT$ gate, so its output is $Y_2 = \overline{Y_1 \cdot Y_1} = \overline{Y_1}$.
Substituting $Y_1$, we get $Y_2 = \overline{\overline{A+B}} = A+B$.
The final output $y$ passes through a $NOT$ gate, so $y = \overline{Y_2} = \overline{A+B}$.
This is the Boolean expression for a $NOR$ gate.
Therefore, the circuit is equivalent to a $NOR$ gate.

The logic gate equivalent to the circuit shown in the figure is

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