Let $a,b,c,d=\tan^2\frac{\pi}{16},$ $\tan^2\frac{3\pi}{16},$ $\tan^2\frac{5\pi}{16},$ $\tan^2\frac{7\pi}{16};$ the sum factors as $(a+d)(b+c)$

$\tan\frac{7\pi}{16}=\cot\frac{\pi}{16},$ and $\tan^2x+\cot^2x=\dfrac{4}{\sin^2 2x}-2$

With $a=\tan^2\tfrac{\pi}{16},\$ $b=\tan^2\tfrac{3\pi}{16},\$ $c=\tan^2\tfrac{5\pi}{16},\$ $d=\tan^2\tfrac{7\pi}{16},$ the expression is $ab+ac+bd+cd$ $=(a+d)(b+c).$

Since $\tfrac{7\pi}{16}=\tfrac{\pi}{2}-\tfrac{\pi}{16},$ we have $d=\cot^2\tfrac{\pi}{16},$ so $a+d=\tan^2\tfrac{\pi}{16}+\cot^2\tfrac{\pi}{16}$ $=\tfrac{4}{\sin^2(\frac{\pi}{8})}-2$ $=14+8\sqrt2.$ Likewise $b+c=\tfrac{4}{\sin^2(\frac{3\pi}{8})}-2$ $=14-8\sqrt2.$ Their product is $14^2-(8\sqrt2)^2=196-128=68.$

Thus, the correct answer is B.