If $X$ and $Y$ are random variables taking on values $x_1, x_2, \ldots, x_m,$ and
$y_1, y_2, \ldots, y_n,$ respectively, then the joint probabilities $P\{X = x_i, Y = y_j\}$ (meaning the probability that $X$ has value $x_i$ and simultaneously $Y$ has value $y_j$) can be displayed as an array (a matrix, if you like)
with $m$ columns and $n$ rows:
$$\scriptstyle\begin{array}{c|c|c|c|c|c}
& x_1 & \cdots & x_i & \cdots & x_m&\\\hline
y_1& P\{X = x_1, Y = y_1\}& \cdots&P\{X = x_i, Y = y_1\}&\cdots
&P\{X = x_m, Y = y_1\}&P\{Y=y_1\}\\\hline
\vdots & \vdots&\ddots&\vdots&\ddots&\vdots&\vdots\\\hline
y_j&P\{X = x_1, Y = y_j\}&\ddots&P\{X = x_i, Y = y_j\}
&\ddots&P\{X = x_m, Y = y_j\}&P\{Y=y_j\}\\\hline
\vdots&\vdots&\ddots&\vdots&\ddots&\vdots&\vdots\\\hline
y_n&P\{X = x_1, Y = y_m\}&\cdots&P\{X = x_i, Y = y_m\}&\cdots
&P\{X = x_m, Y = y_n\}&P\{Y=y_n\}\\\hline
&P\{X=x-1\}&\cdots&P\{X = x_i\}&\cdots&P\{X = x_m\}& 1
\end{array}$$
Notice that in addition to the labels $x_1, x_2, \ldots, x_m,$ and
$y_1, y_2, \ldots, y_n,$ along the top and on the left
respectively, I have added a rightmost ($(m+1)$-th) column
and a bottommost ($(n+1)$-th) row. The entries here are
the row sums and column sums respectively: the sum of
the entries in the $i$-th column is
$$P\{X=x_i, y=y_1\}+\cdots+P\{X=x_i, Y=y_j\}+\cdots+P\{X=x_i, Y=y_n\}
= \sum_{j=1}^n P\{X=x_i,Y=y_j\}$$
which is just the marginal probability $P\{X = x_i\}$.
(The reason for calling this a marginal probability is that
it is written down in the margins of the array: here, marginal
is not used in the other meaning of the word as something that
barely meets the standards or barely gets a passing grade).
Note that the formula $\displaystyle P\{X = x_i\} = \sum_{j=1}^n P\{X=x_i,Y=y_j\}$
is just your P(A) = ∑ p(A,B) where, as noted by DeepakML, the sum is over events $B$ that are mutually exclusive: $Y$ cannot simultaneously have
two different values $Y_i$ and $Y_k$, and exhaustive: $Y$ must take
on one of the values $y_1, y_2, \ldots, y_n$. In other words, the
mutual exclusivity does not apply to $A$ and $B$ as you appear to think.
