Let set $\mathbb{A}$ contain all those permutations which do not have property $P$, and set $\mathbb{B}$ be the set contain those permutations each of which contains $\underline{exactly\ one}$ pair of $(x_i,\ x_{i+1})$ satisfying $\mid x_i − x_{i+1}\mid = n$. Then the number of elements in $\mathbb{B}$ must be less than the number of all permutations with property $P$ because there exist permutations having more than one pair satisfying the condition. If we can show that the number of elements in $\mathbb{A}$ is no more than that in $\mathbb{B}$, then the to-be-proved claim will hold.

Firstly, we are going to show that every element in $\mathbb{A}$ has a corresponding element in $\mathbb{B}$. Let a set $\{x_1,\ x_2,\ \cdots,\ x_n\}\in\mathbb{A}$ and $x_k$ be the unique number that satisfies $\mid x_1 - x_k\mid = n$, then $k > 2$. Define mapping $\mathcal{f}$ as following $$\{x_1,\ x_2,\ \cdots,\ x_n\} \Rightarrow \{x_2,\ \cdots,\ x_{k-1},\ x_1,\ x_k,\ \cdots, x_n\}$$

i.e. moving $x_1$ to the position before $x_k$ and shifting each of $x_2$, $\cdots$, $x_{k-1}$ one position forward. Then the mapped element must be in $\mathbb{B}$ because of the pair $(x_1,\ x_k)$ meets the condition.

Meanwhile, different elements in $\mathbb{A}$ must be mapped to different elements in $\mathbb{B}$ by $\mathcal{f}$. This is because every element in $\mathbb{B}$ has only one pair of $(x_i,\ x_{i+1})$ satisfying the condition $\mid x_i - x_{i+1}\mid = n$. Therefore, the result of moving $x_i$ to the first position of the sequence and each of $x_1$, $\cdots$, $x_{i-1}$ to the next position, is deterministic and unique.

Combining these two conclusions shows that the number of elements in $\mathbb{A}$ cannot be more than that in $\mathbb{B}$. Then, the original claim follows.