Power set, Why is a right? 
Can someone tell me why the first choice is the correct answer?
 A: Power set is the set containing all subsets of S.
{a,b} is a subset of S. So, it is an element of P.
{{a}} is also a subset of S. So, it is an element of P.
similarly, the rest follow.
A: The power set of a set is the set of all the subsets of a set.
We have that 
$$S = \{a, b, \{a\}\}$$
Therefore, the power set would be:
$$P(S) = \{\{\}, \{a\}, \{b\}, \{\{a\}\}, \{a, b\}, \{a, \{a\}\}, \{b, \{a\}\}, \{a, b, \{a\}\}\}$$
Therefore, looking at the answers, we can deduce that (a). is the answer.
A: i) $\{b\} \in S$.  This is false the set $\{b\}$ is not a member of S.  The members of S are:$a$, $b$, and the set $\{a\}$
ii) $\{a\} \subset P(S)$. This is subtle but that is false.  $\{a\} \subset S$ so $\{a\} \in P(S)$.  But P(S) is the set of all subsets. The only elements it has will be sets.  Any subset of P(S) will be a set containing subsets.  {a} is not a set of sets.  It is simply a set with an element.
Perhaps it will help to list the elements of P(S).  P(S) = {$\emptyset$, {a}, {b}, {a}, {a,b}, {a,{a}}, {b,{a}}, {a,b,{a}}}.  As you can see $a \not \in P(S)$  so $\{a\} \not \subset P(S)$.
iii){a,b} $\in P(S)$.  True.  {a,b} $\subset S$ so $\{a,b\} \in P(S)$.
The entire trick of these is to understand the difference between being an element of a set and being in the set versus being a subset of a set, and being comfortable with the idea of a "set of sets" having sets as its elements.
iv) $\{a,b\}\in S$.  False.  Again {a,b} is not an element of S.
v){{a}} $\in P(s)$.  True.  {{a}} $\subset S$ so $\{\{a\}\} \in P(S)$.
vi){a.{a}} $\in P(S)$.  True $\{a,\{a\}\} \subset S$ so $\{a,\{a\}\} \in P(S)$.
