The given circuit works as:

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(C) Let the inputs be $A$ and $B$. The first two $NOR$ gates act as $NOT$ gates because their inputs are shorted.$P = \overline{A+A} = \overline{A}$$Q

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$NAND$ gate

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(C) Let the inputs be $A$ and $B$. The first two $NOR$ gates act as $NOT$ gates because their inputs are shorted.$P = \overline{A+A} = \overline{A}$$Q = \overline{B+B} = \overline{B}$These are inputs to the third $NOR$ gate, so the output $R$ is:$R = \overline{P+Q} = \overline{\overline{A} + \overline{B}}$Using De Morgan's Law, $\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$This $R$ is the input to the final $NOR$ gate, which acts as a $NOT$ gate:$S = \overline{R+R} = \overline{R} = \overline{A \cdot B}$Since the final output is $\overline{A \cdot B}$, the circuit functions as a $NAND$ gate.

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