(N/A) and $B$ are the inputs and $Y$ is the output of the given circuit. The left part of the circuit is a $NOR$ gate,and the right part is a $NOT$ gate.
The output of the $NOR$ gate is $\overline{A+B}$.
This output serves as the input for the $NOT$ gate. The output of the $NOT$ gate is $\overline{\overline{A+B}} = A+B$.
Therefore,$Y = A+B$,which is the Boolean expression for an $OR$ gate. Hence,this circuit functions as an $OR$ gate.
$(b)$ $A$ and $B$ are the inputs and $Y$ is the output. The inputs $A$ and $B$ first pass through two $NOT$ gates,producing $\overline{A}$ and $\overline{B}$ respectively.
These are then fed as inputs to a $NOR$ gate.
The output of the $NOR$ gate is $Y = \overline{\overline{A} + \overline{B}}$.
Using De Morgan's theorem,$\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$.
Therefore,$Y = A \cdot B$,which is the Boolean expression for an $AND$ gate. Hence,this circuit functions as an $AND$ gate.