Denote $P_{n+1} = P_1$, $X_{n+1} = X_1$, $A_{n+1} = A_1$.

Lemma: Let $Q$ be a point inside $A_1A_2\cdots A_n$. Then $Q$ must lie inside one of the circumscribed circles of the triangles $X_1A_2X_2, \cdots, X_nA_1X_1$.

Proof: If $Q$ lies in one of the triangles $X_1A_2X_2, \cdots, X_nA_1X_1$, then the conclusion clearly holds. Otherwise $Q$ lies inside the polygon $A_1A_2\cdots A_n$ (as in Figure 1). Then
$$\begin{aligned} & (\angle X_1 A_2 X_2 + \angle X_1 Q X_2) + \cdots + (\angle X_n A_1 X_1 + \angle X_n Q X_1) \\ &= (\angle X_1 A_1 X_2 + \cdots + \angle X_n A_1 X_1) + \cdots + (\angle X_1 Q X_2 + \cdots + \angle X_n Q X_1) \\ &= (n-2)\pi + 2\pi = n\pi, \end{aligned}$$
therefore there exists an index $i$ such that
$$\angle X_i A_{i+1} X_{i+1} + \angle X_i Q X_{i+1} \ge \frac{n\pi}{n} = \pi.$$
Since the quadrilateral $QX_iA_{i+1}X_{i+1}$ is convex, this means that $Q$ lies inside the circumscribed circle of $\Delta X_iA_{i+1}X_{i+1}$.

Applying the above lemma, $P$ lies inside the circumscribed circle of some $\Delta X_iA_{i+1}X_{i+1}$.

Consider respectively the circumscribed circles $\omega$ and $\Omega$ of $\Delta P_iA_{i+1}P_{i+1}$ and $\Delta X_iA_{i+1}X_{i+1}$ (as in Figure 2); let $r$ and $R$ be their radii respectively. Then we obtain
$$2r = A_{i+1}P \le 2R \text{ (since $P$ lies inside $\Omega$)},$$
hence
$$P_i P_{i+1} = 2r \sin \angle P_i A_{i+1} P_{i+1} \le 2R \sin \angle X_i A_{i+1} X_{i+1} = X_i X_{i+1}.$$

Fig. 1

Fig. 2