The scenario below will help me understand how to apply the law of total probability to nonbinary examples. It is inspired by an exercise from a master's program in AI using Bayesian Networks I'm currently under. The scenario is fictional and it was made purely to show how to use the theorem with a more complicated example. The law of total probability is a requirement for understanding the probability of dependent variables and events.
Let $A$ and $B$ be independent identically distributed random variables. $A$ and $B$ can assume values from $[0, 1, 2]$ with probabilities $[0.25, 0.25, 0.50]$ respectively. Now, let event $E$ be dependent on $A$ and $B$, so the probability of $E$ being $X$, $Y$ or $Z$ can be summarized as follows:
\begin{array}{c|lll} E(B-A) & X & Y & Z \\\hline 0 & 0.15 & 0.15 & 0.70 \\ 1 & 0.25 & 0.40 & 0.35 \\ 2 & 0.10 & 0.65 & 0.25 \end{array}
How to compute $P(E(B-A)=X)$, that is the probability of $E$ being $X$?
$\newcommand{\P}{\mathbb{P}}$From what I can see, $E$ is a random variable, with the rows of your table giving us the conditional distribution of $E$ given the value of $|B-A|$ (I am assuming it is $|B-A|$ because the table only has non-negative values but $B-A$ could he negative). So the first column of your table is telling us $\P(E = X\mid |B-A|)$. Hence to get $\P(E=X)$, you should use the law of total probability, which tells us that $$\P(E=X) = \sum_{k=0}^2 \color{blue}{\P(E=X\mid |B-A| =k)}\color{red}{\P(|B-A|=k)}.$$
The $\color{blue}{\P(E=X\mid |B-A|=k)}$ are in the first column of your table. So, compute the probabilities $\color{red}{\P(|B-A| = k)}$ for $k=0,1,2$ (make sure you know how to do this!) and you will have everything you need to work out your desired probability.