Identify the gate and match A, B, Y in the bracket to check.

The output of the first NAND gate

${Y}^{'} = \overset{¯}{A \cdot B}$

The output of the second NAND gate

$Y = \overset{¯}{\left(\overset{¯}{A \cdot B}\right) \cdot \left(\overset{¯}{A \cdot B}\right)} ⇒ Y = \left(\overset{¯}{\overset{¯}{A \cdot B}}\right) + \left(\overset{¯}{\overset{¯}{A \cdot B}}\right) ⇒ Y = A \cdot B$

So, the equivalent circuit will be an AND gate.

$AND ( A = 1, B = 1, Y = 1 )$