The truth table for this given circuit is:

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(B) The given circuit consists of two $AND$ gates and one $OR$ gate. The inputs to the first $AND$ gate are $A$ and $B$,so its output is $A \cdot B$.T

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$A$$B$$Y$$0$$0$$0$$0$$1$$1$$1$$0$$0$$1$$1$$1$

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(B) The given circuit consists of two $AND$ gates and one $OR$ gate. The inputs to the first $AND$ gate are $A$ and $B$,so its output is $A \cdot B$.The input $A$ passes through a $NOT$ gate,becoming $\overline{A}$. This $\overline{A}$ and input $B$ are fed into the second $AND$ gate,so its output is $\overline{A} \cdot B$.These two outputs are then fed into an $OR$ gate,giving the final output $Y = (A \cdot B) + (\overline{A} \cdot B)$.Using Boolean algebra: $Y = (A + \overline{A}) \cdot B$.Since $A + \overline{A} = 1$,we get $Y = 1 \cdot B = B$.Thus,the output $Y$ is equal to input $B$. Checking the truth table for $Y = B$:If $A=0, B=0$,then $Y=0$.If $A=0, B=1$,then $Y=1$.If $A=1, B=0$,then $Y=0$.If $A=1, B=1$,then $Y=1$.This matches option $B$.

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$A$	$B$	$Y$
$0$	$0$	$1$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$0$

$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$1$
$1$	$0$	$0$
$1$	$1$	$1$

$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$0$
$1$	$0$	$0$
$1$	$1$	$1$

$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$0$