Using Vieta's formulas, we have:
\begin{align*}a+b+c+d &= 18,\\ ab+ac+ad+bc+bd+cd &= k,\\ abc+abd+acd+bcd &=-200,\\ abcd &=-1984.\\ \end{align*}
From the last of these equations, we see that $cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62$ . Thus, the second equation becomes $-32+ac+ad+bc+bd+62=k$ , and so $ac+ad+bc+bd=k-30$ . The key insight is now to factor the left-hand side as a product of two binomials: $(a+b)(c+d)=k-30$ , so that we now only need to determine $a+b$ and $c+d$ rather than all four of $a,b,c,d$ .
Let $p=a+b$ and $q=c+d$ . Plugging our known values for $ab$ and $cd$ into the third Vieta equation, $-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)$ , we have $-200 = -32(c+d) + 62(a+b) = 62p-32q$ . Moreover, the first Vieta equation, $a+b+c+d=18$ , gives $p+q=18$ . Thus we have two linear equations in $p$ and $q$ , which we solve to obtain $p=4$ and $q=14$ .
Therefore, we have $(\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30$ , yielding $k=4\cdot 14+30 = \boxed{86}$ .