$\color{maroon}{a+7b+3c+5d=16}$
$\color{maroon}{2a+6b+4c+8d=16}$
Or, $(a+a)+(7b-b)+(3c+c)+ (5d+3d) = 16$
Or, $(a+7b+3c+5d)+(a-b+c+3d) = 16$
Or, $a-b+c+3d = 0 \qquad\qquad ∵\big[\color{blue}{ a+7b+3c+5d=16}\big]$ ------------- 1)
$\color{maroon}{8a+4b+6c+2d = -16}$
$\color{maroon}{5a+3b+7c+d = -16}$
Or, $(5a+3a)+(3b+b)+(7c-c)+(d+d) = -16$
Or, $(5a+3b+7c+d)+(3a+b-c+d) = -16$
Or, $3a+b-c+d = 0 \qquad \qquad \big[∵\color{blue}{5a+3b+7c+d = -16}\big]$ ------------- 2)
Adding 1) & 2)
$3a+b-c+d=0\\a-b+c+3d=0$
Or, $4a +4d = 0$
Or, $a+d = 0$
Now, we've to find $\color{maroon}{(a+d)(b+c) = ?}$
$(a+d)(b+c) = (0)\times(b+c) \qquad \qquad ∵ \big[\color{blue}{a+d=0}\big]$
∴$\color{red}{(a+d)(b+c) = 0}$
$\color{green}{\text{Hence, option }} \color{orange}{ B)} \color{green}{ \text{ is the right answer}}$.