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The given circuit (two NOT gates feeding into a NOR gate, inputs A and B) is equivalent to: [add image]
A
[add image: OR gate symbol, inputs A,B, output Y]
B
[add image: NAND gate symbol, inputs A,B, output Y]
C
[add image: OR gate symbol variant]
D
[add image: AND gate symbol, inputs A,B, output Y]
Detailed Solution
$Y=\overline{\overline{A}+\overline{B}}=\overline{\overline{A}}\cdot\overline{\overline{B}}=A\cdot B$. This is the truth table of an AND gate, so the given combination of gates is equivalent to a single AND gate.