Given:

$A=\left[\begin{array}{ccc}a& 3& 5\\ 5& -1& 3\\ 2& 3& -4\end{array}\right]$ 

$B=\left[\begin{array}{ccc}b& 1& 4\\ 4& c& 1\\ -3& 1& d\end{array}\right]$

Since, the trace of $A$ is $-4$, so

$a-1-4=-4$

$\Rightarrow a=1$

Also,

$AB=\left[\begin{array}{ccc}-1& 0& 17\\ -3& 10& 25\\ 28& -8& 3\end{array}\right]$

$\Rightarrow \left[\begin{array}{ccc}1& 3& 5\\ 5& -1& 3\\ 2& 3& -4\end{array}\right]\left[\begin{array}{ccc}b& 1& 4\\ 4& c& 1\\ -3& 1& d\end{array}\right]=\left[\begin{array}{ccc}-1& 0& 17\\ -3& 10& 25\\ 28& -8& 3\end{array}\right]$

$\Rightarrow \left[\begin{array}{ccc}b-3& 6+3c& 7+5d\\ 5b-13& 8-c& 19+3d\\ 2b+24& 3c-2& 11-4d\end{array}\right]=\left[\begin{array}{ccc}-1& 0& 17\\ -3& 10& 25\\ 28& -8& 3\end{array}\right]$

$\therefore b-3=-1\Rightarrow b=2$

$6+3c=0\Rightarrow c=-2$

$7+5d=17\Rightarrow d=2$

Hence, $a+b+c+d=1+2-2+2=3$