jmc

Let's perform the addition one step at a time. The first step is adding $c$ and $a$ in the right column. Since $c$ and $a$ can't both be 0 and $c+a$ is at most $9+8=17$, we know that $c+a=10$. The one carries over.

The second step is adding $b$ and $c$ in the middle column. Similarly, we know that $b+c+1=10$ (the one is from the carrying over), so $b+c=9$. The one carries over.

The third step is adding $a$ and $d$ in the left column. Similarly, we know that $a+d+1=10$ so $a+d=9$.

Thus, we have the three equations $$\begin{array}{rl}a+c& =10\\ b+c& =9\\ a+d& =9\end{array}$$ Adding the last two equations gives $b+c+a+d=9+9=18$, so our answer is $\boxed{18}$. This corresponds to $(a,b,c,d)\Rightarrow (4,3,6,5)$.