Let $k_i$ be the number of sets that $i$ is contained in. Then $P_i \cap P_j = \{a\}$ for exactly $\binom{k_a}{2}$ pairs of subsets. Note that if $P_i \cap P_j$ is nonempty then it is a one element set, so summing over all $a$ we get that $P_i \cap P_j$ is nonempty for number of pairs equal to \[ \binom{k_1}{2} + \binom{k_2}{2} + \cdots + \binom{k_n}{2}. \] We also have $k_1 + k_2 + \cdots + k_n = 2n$. Using AM-GM/Cauchy/whatever we can find the minimum of this expression is $n$ with equality iff $k_1 = k_2 = \cdots = k_n = 2$. On the other hand, each such pair $P_i \cap P_j \neq \emptyset$ gives us a $k$ with $P_k = \{i,j\}$. There are only $n$ subsets so there are at most $n$ such $k$. So in fact we must have equality above, and $k_1 = k_2 = \cdots = k_n = 2$. This immediately implies the statement we want to prove.