Is $\emptyset$ considered as a powerset by itself? For example, $X = \{\emptyset, a, \{b\}\}$. Find the power set of $X$.
As far as I believe everyone understand, a power set of something means to display whatever element is within the set itself.
So for instance, $\{b\}$ contain the element $b$ inside while $a$ is just an element by itself. So in order to conclude a power set of a $\emptyset$, a set that contains element $b$, and a element $a$ itself.
Isn't it right to right the power set of $X$ as 
$$ P\{\emptyset, a, \{b\}\} = \{\emptyset, \{a\}, \{b\}, \{a,b\}\}?$$
However, is $\{\emptyset,a\}$ is logically similar to the $\{a, \emptyset\}$? Or $\emptyset$ should just remain as a $\emptyset$ and should not be placed with other elements in the set like the answer above?
 A: Your set $X$ has three distinct elements (provided $a\ne\emptyset$ and $a\ne\{b\}$), hence your power set should have eight elements. 
You may find it simper to proceed this way: First, assume $X=\{u,v,w\}$ and write down the power set of $X$ (it should contain eight elements).
Then, replace each occurrence of $u$ with $\emptyset$, each occurrence of $v$ with $a$, and each occurrence of $w$ with $\{b\}$. 
Doing so, you should notice that not only you get more elements than you suspected, but also that $\{b\}$ and $\{a,b\}$ are not among them.
Regarding your closing sentence: It is always the case that $\{x,y\}=\{y,x\}$.
A: Having an element $ \emptyset \in A $ is not tautological. For example, $ \emptyset \notin \emptyset $. As such, the power set of $ X $, has exactly $ 8 = |\wp(X)| $ elements (more on this below).
The powerset of $ \emptyset $ is not $ \emptyset $, instead $ \wp(\emptyset) = \{ \emptyset \} $, since it follows from the definition of powersets, that $$ x \in \wp(A) \Leftrightarrow x \subseteq A $$
Since $ \emptyset \subseteq A $, $ \emptyset \in \wp(A) $. Therefore having $ \wp(\emptyset) = \emptyset $ would be contradictive.

Isn't it right to right the power set of $X$ as 
$$ P\{\emptyset, a, \{b\}\} = \{\emptyset, \{a\}, \{b\}, \{a,b\}\}?$$
However, is $\{\emptyset,a\}$ is logically similar to the $\{a, \emptyset\}$? Or $\emptyset$ should just remain as a $\emptyset$ and should not be placed with other elements in the set like the answer above?.

Well, $ \emptyset \in X $ is no different from other elements in $X$, so of course it is present in the subsets. Take $ \{\emptyset, a \} \subseteq X $; it follows from the definition of powersets that this is in fact in $ \wp(X) $.
Sets are sets, collections of elements: either $ a \in A $ or $ a \notin A $, the order doesn't matter.
Now, there is another mistake in your formula: you assume that $ b \in X $, which isn't the case, $ b $ is in a set ($ \{b\} $), which is in $ X $.
So, in conclusion $$
\wp(X) = \{ \emptyset, \{ \emptyset \}, \{ a \}, \{ \{b\}\}, \{a, \{b\}\}, \{\emptyset, a\}, \{\emptyset, \{b\}\}, \{\emptyset, a, \{b\}\}\}.
$$
A: $\{\emptyset\}\neq\{\}=\emptyset$
If a set containing the empty set were the empty set, then your proposition would be true.  But a set containing the empty set is not the empty set.  This is the specific purpose of the axiom of regularity.
$P(X)=\{\{\},\{\emptyset\},\{a\},\{\{b\}\},\{\emptyset,a\},\{\emptyset,\{b\}\},\{a,\{b\}\},\{\emptyset,a,\{b\}\}\}$
$P(X)=\{\emptyset,\{\emptyset\},\{a\},\{\{b\}\},\{\emptyset,a\},\{\emptyset,\{b\}\},\{a,\{b\}\},\{\emptyset,a,\{b\}\}\}$
