The output of $2\times 4$ decoder is act as input for $4\times 1$ multiplexer. So the output function of multiplexer is:
$F(A,B,C,D)=\bar D\bar CI_0+\bar DCI_1+D\bar CI_2+DCI_3$
$F(A,B,C,D)=\bar D\bar C\bar AB+\bar DC\bar A\bar B+D\bar CAB+DCA\bar B$
$F(A,B,C,D)=\bar AB\bar C\bar D+\bar A\bar BC\bar D+AB\bar CD+A\bar BCD$
$F(A,B,C,D)=\bar A\bar D(B\bar C+\bar BC)+AD(B\bar C+\bar BC)$
$F(A,B,C,D)=(B\bar C+\bar BC)(\bar A\bar D+AD)$
$F(A,B,C,D)=(B\oplus C)(A\odot D)$
So option (C) is correct.