We don't know actual numeric values for $a, b$, or $d$, but we know that $a$ and $b$ are negative, while $d$ is positive. And since the tick marks are equally spaced, we know that $a=2 b$, and that $d=-a$. Sometimes it'll be easier to solve these mathematically, but other times we can solve them graphically.
Since we need to select all true statements, we need to consider each one in turn.
(A) $\(a b+a d>0\)$

At a quick glance, we know that $a b$ is the product of two negative numbers, so it's positive, while $a d$ is the product of a negative and a positive number, so it's negative. But which of those two terms has the greater absolute value? That will determine whether or not the sum is positive.
One approach is to simplify: $\(a b+a d=a(b +d)\)$. Since $\(d=-a=-2 b\)$, we know that $\(b +d=b-2 b=-b\)$. Since $b$ is negative, $-b$ is positive; since $a$ is negative also, and $\(a(-b)\)$ is negative, so $\(a b+a d<0\)$. This statement is false.

Another approach is just to eyeball it: the absolute value of $a b$ is going to be smaller than the absolute value of $a d$, since $b$ is a smaller number (in absolute value) than $d$. So $a b$, the positive term, is smaller than $a d$, the negative term, and the sum will be negative.
(B) $\(b-a=d\)$

Graphically, we can see $b-a$ yields a result that is one tick mark on the positive side of $O$ : since $a$ is two tick marks on the negative side of O , subtracting a will yield a result two tick marks to the right of $b$. Since $d$ is two tick marks away from 0 , though, this statement must be false.

Mathematically, since $a=2 b, b-a=b-2 b =-b$, but $d=-a=-2 b$, so $b-a \neq d$.
(C) $\((a b)(b d)(a d)>0\)$

This term is equivalent to $(a b d)^2$. Any number squared is greater than 0 , so this statement is true.
(D) $\(\frac{a}{d}+\frac{b}{d}<0\)$

Dividing a negative number by a positive number yields a negative number, so each of these terms are negative and so is their sum. This statement is true.

Choices (C) and (D) must be true.
An alternative method is to Pick Numbers. A scan of the answer choices tells you that you must have numbers that are easy to divide. Let's make $a=-4, b=-2$, and $d=4$. Testing each answer choice with these values will again establish that only choices (C) and (D) must be true.