For the circuit shown above, equivalent GATE is :

$$ Y_1 = \overline{A \cdot B} $$
$$ Y_2 = \overline{A + B} $$
$$ Y = \overline{Y_1 \cdot Y_2} $$
$$ Y = \overline{ (\overline{A \cdot B}) \cdot (\overline{A + B}) } $$

Apply De Morgan's Law :
$$ Y = \overline{ (\overline{A \cdot B}) } + \overline{ (\overline{A + B}) } $$

The double inversions cancel out:
$$ Y = (A \cdot B) + (A + B) $$

$$ Y = A \cdot B + A + B $$
$$ Y = A \cdot (B + 1) + B $$

Since anything ORed with 1 results in 1 :
$$ Y = A \cdot 1 + B $$
$$ Y = A + B $$

Therefore, the equivalent gate for the entire circuit is an OR gate.