The truth table for this given circuit is:

$$\text{Top AND gate inputs} = A \text{ and } B \implies \text{Output} = A \cdot B$$

$$\text{Bottom AND gate inputs} = \bar{A} \text{ and } B \implies \text{Output} = \bar{A} \cdot B$$

$$Y = (A \cdot B) + (\bar{A} \cdot B)$$

$$Y = (A + \bar{A}) \cdot B = 1 \cdot B \implies Y = B$$ 

Hence, $$Y$$ is independent of input $$A$$

$$A = 0,\ B = 0 \implies Y = 0$$ 

$$A = 0,\ B = 1 \implies Y = 1$$

$$A = 1,\ B = 0 \implies Y = 0$$

$$A = 1,\ B = 1 \implies Y = 1$$