The notation $\mathscr P(S_n)$ means the power set of $S_n$, which is the set of all subsets of $S_n$. For example,
$\mathscr P(S_2)=\{\emptyset,\{a_1\},\{a_2\},\{a_1,a_2\}\}.$
This problem is looking at the relationship between the power set of a set and the same set with one extra element, and the answer shows that the new set $\mathscr P(S_{n+1})$ is actually twice as big as $\mathscr P(S_n)$, even though only one new element was added to $S_n$. In response to your question, $S_{n+1}=S_n\cup\{a_{n+1}\}$ is true, but $\mathscr P(S_{n+1})=\mathscr P(S_n)\cup\{a_{n+1}\}$ is not. To continue with the example for $S_2$, let's look at $\mathscr P(S_3)$:
$\mathscr P(S_3)=\{\emptyset,\{a_1\},\{a_2\},\{a_1,a_2\},\{a_3\},\{a_1,a_3\},\{a_2,a_3\},\{a_1,a_2,a_3\}\}.$
Notice that it can be seen as two separate pieces: the first four elements are the same as $\mathscr P(S_2)$, but there are four new elements, each of which contain the new element $a_3$. In fact, you can get these four elements by taking each element $A\in\mathscr P(S_2)$ and adding $a_3$ to it, to get $A\cup\{a_3\}$. The set of all of these, taken together, is $\{A\cup\{a_3\}:A\in\mathscr P(S_2)\}$, and putting that together with the original set $\mathscr P(S_2)$, we get
$\mathscr P(S_3)=\mathscr P(S_2)\cup\{A\cup\{a_3\}:A\in\mathscr P(S_2)\}.$
Generalizing the pattern, we arrive at the desired result:
$\mathscr P(S_{n+1})=\mathscr P(S_n)\cup\{A\cup\{a_{n+1}\}:A\in\mathscr P(S_n)\}.$