The logical expression can be written as:
$\left[\overline{(A+B)}+C\right].\left[(C+D)\right]$
Applying De-morgans' law:
$\left[(\bar A.\bar B)+C\right].(C+D)$
$\bar A\bar B C+\bar A\bar BD+C+CD$
taking $C$ as common here;
$C(\bar A\bar B+1+D)+\bar A\bar BD$
$\because (1+x=1)$
$\therefore \bar A\bar BD+C$
Note: none of the options are correct here.