As $x$ is connected to $I0$ & $I1$, $y$ connected to $I2,$ $y'$ connected to $I3$ & $A1,$ $z$ connected to $A0$ and $z'$ connected to ENABLE $(EN),$
$f= ( {\overline{A1}}.{\overline{A0}}.I0+ {\overline{A1}}.A0.I1 +A1.{\overline{A0}}.I2 + A1.A0.I3 ).EN$
$\implies f = (xyz' +xyz + y'z'y + zy')z'$
$\qquad= (xyz' +xyz + zy')z' =xyz'$
Correct Answer: $A$