The concept to be used: For all $a$ and $b$, we have: $|a| + |b| ≥ |a + b|$

For example:

#1. $a = 2, b = 3 : |a| + |b| = |2| + |3| = 2 + 3 = 5$ and $|a + b| = |2 + 3| = 5$ => $|a| + |b| = |a + b|$

#2. $a = -2, b = -3 : |a| + |b| = |-2| + |-3| = 2 + 3 = 5$ and $|a + b| = |-2 - 3| = 5$ => $|a| + |b| = |a + b|$

#3. $a = -2, b = 3 : |a| + |b| = |-2| + |3| = 2 + 3 = 5$ and $|a + b| = |-2 + 3| = 1$ => $|a| + |b| > |a + b|$

Note:
# $|a| + |b| = |a + b|$ if $a$ and $b$ are of the same sign
# $|a| + |b| > |a + b|$ if $a$ and $b$ are of opposite signs

We have been given that: $|a| + |b| = |a + b|$ => $a$ and $b$ are of the same sign

Thus, $a*b > 0$

Answer B