Set of elements that don't belong to any power set of these elements [Chartrand P242 10.22 2nd Ed = 10.30 3rd Ed] 
How and why is the answer $B = A_d$ ? 
I thought : Because $A := \{a, b, c\} \neq \{d, e, f, g, h\}$,
thus $B = \bigcup_{i\in \{d, e, f, g, h\}} A_i$.

Supplementary dated Dec 12 2013:
By the definition of $B$, we consider $x \in A = \{a, b, c\} $. Thus, are $A_d, A_e, ..., A_h$ really redundant and immaterial to the process of determining $B$? I'm sensing that I should've realised this before doing any work; user rewritten writes that "the rest of the $A_x$ do not enter..." and user Potato analyses only $A_a$, $A_b$, and $A_c$.
 A: I am not quite sure how you are obtaining your answer. I will try to describe how I thought about this. Hopefully it is helpful.
We want the determine $B$. The set $B$ is defined as the set of all elements in $A$ satisfying a certain property: $x$ is not contained in $A_x$. Let us examine each element of $A$ in turn and see if it is in $B$.
Since $A_a$ is null, $a$ is not in $A_a$, so $a$ is in $B$ (because it satisfies the defining property).
Since $A_b=A=\{a,b,c\}$, b is in $A_b$, so it is not in $B$ (because it does not satisfy the defining property).
Since $A_c=\{a,b\}$, $c$ is not in $A_c$, so $c$ is in $B$.
So $a$ and $c$ work, and we obtain $B=\{a,c\}$.
In contrast, your proposed $B$ is $\{a,b,c\}$  (since the sets you are unioning contain all of these elements). 
The fact that the answer ends up being $A_d$ is entirely coincidental. 
Regarding your edit, you are correct that the rest of the $A_x$ are irrelevant to the process of determining $B$. 
A: Being @potato's answer completely correct, I try to clear up the fundamental misunderstanding.
$A$ is the set $\{a,b,c\}$. Let's define a function $g:A\to\mathcal{P}(A)$ such that:
$$
g(a) = A_a = \emptyset \\
g(b) = A_b = A \\
g(c) = A_c = \{a,b\}
$$
The rest of the $A_x$ do not enter at all in this definition.
Now you need to find B, and finally show that B is not in the image of $g$. It's completely irrelevant whether $B$ is $A_d$ or $\omega$ or anything else. The point is that it's different from $g(a)$, $g(b)$ and $g(c)$, so it's not in the image of $g$.
