$F(A,B,C,D)=\overline{\overline {BD}.(\overline{\bar BC})}$
$F(A,B,C,D)=\overline {\overline{BD}}+\overline{\overline{\bar BC}}$
$F(A,B,C,D)=BD+\bar BC$
Minterms are: $\sum m(2,3,5,7,10,11,13,15)$
So total $8$ minterms are present in the given function.