Input for $4\times1$ multiplexer is:
$I_0=\bar C\bar D+C\bar D=\bar D(C+\bar C)=\bar D$
$I_1=I_2=0$
$I_3=\bar C D+C D= D(C+\bar C)= D$
$\therefore F=\bar A\bar BI_0+\bar A BI_1+ A\bar BI_2+ A BI_3$
$F=\bar A\bar B\bar D+\bar A B.0+ A\bar B.0+ A BD$
$F=\bar A\bar B\bar D+ A BD$
Option $(D)$ is correct.