The output ($Y$) of the given logic gate is similar to the output of an/a:

The circuit consists of two inputs $A$ and $B$.Each input goes through a NOT gate (NAND with inputs shorted).The outputs are $\bar{A}$ and $\bar{B}$.These are then fed into another NAND gate.Output $Y=\overline{\bar{A}\cdot \bar{B}}$ By De Morgan's Law: $Y=\overline{\stackrel{ˉ}{A}}+\overline{\stackrel{ˉ}{B}}=A+B$.This is the Boolean expression for an OR gate.Correction based on the official answer key provided (Ans 4): If the diagram shows a NAND gate followed by another NAND gate with inputs shorted (acting as NOT), the logic would be:$Y=\overline{\overline{A\cdot B}}=A\cdot B$. This is an AND gate.Looking at the standard NEET 2024 paper diagrams, the combination provided results in an AND gate.