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

Lemma. Let point $Q$ lies inside $A_{1} A_{2} \ldots A_{n}$. Then it is contained in at least one of the circumcircles of triangles $X_{1} A_{2} X_{2}, \ldots, X_{n} A_{1} X_{1}$.

Proof. If $Q$ lies in one of the triangles $X_{1} A_{2} X_{2}, \ldots, X_{n} A_{1} X_{1}$, the claim is obvious. Otherwise $Q$ lies inside the polygon $X_{1} X_{2} \ldots X_{n}$ (see Fig. 1). Then we have
$$\begin{aligned} & \left(\angle X_{1} A_{2} X_{2}+\angle X_{1} Q X_{2}\right)+\cdots+\left(\angle X_{n} A_{1} X_{1}+\angle X_{n} Q X_{1}\right) \\ & \quad=\left(\angle X_{1} A_{1} X_{2}+\cdots+\angle X_{n} A_{1} X_{1}\right)+\left(\angle X_{1} Q X_{2}+\cdots+\angle X_{n} Q X_{1}\right)=(n-2) \pi+2 \pi=n \pi \end{aligned}$$
hence there exists an index $i$ such that $\angle X_{i} A_{i+1} X_{i+1}+\angle X_{i} Q X_{i+1} \geq \frac{\pi n}{n}=\pi$. Since the quadrilateral $Q X_{i} A_{i+1} X_{i+1}$ is convex, this means exactly that $Q$ is contained the circumcircle of $\triangle X_{i} A_{i+1} X_{i+1}$, as desired. $\square$

Now we turn to the solution. Applying lemma, we get that $P$ lies inside the circumcircle of triangle $X_{i} A_{i+1} X_{i+1}$ for some $i$. Consider the circumcircles $\omega$ and $\Omega$ of triangles $P_{i} A_{i+1} P_{i+1}$ and $X_{i} A_{i+1} X_{i+1}$ respectively (see Fig. 2); let $r$ and $R$ be their radii. Then we get $2 r=A_{i+1} P \leq 2 R$ (since $P$ lies inside $\Omega$ ), hence
$$P_{i} P_{i+1}=2 r \sin \angle P_{i} A_{i+1} P_{i+1} \leq 2 R \sin \angle X_{i} A_{i+1} X_{i+1}=X_{i} X_{i+1},$$
QED.

Fig. 1

Fig. 2

As in Solution 1, we assume that all indices of points are considered modulo $n$.

We will prove a bit stronger inequality, namely
$$\max \left\{\frac{X_{1} X_{2}}{P_{1} P_{2}} \cos \alpha_{1}, \ldots, \frac{X_{n} X_{1}}{P_{n} P_{1}} \cos \alpha_{n}\right\} \geq 1,$$
where $\alpha_{i}(1 \leq i \leq n)$ is the angle between lines $X_{i} X_{i+1}$ and $P_{i} P_{i+1}$. We denote $\beta_{i}=\angle A_{i} P_{i} P_{i-1}$ and $\gamma_{i}=\angle A_{i+1} P_{i} P_{i+1}$ for all $1 \leq i \leq n$.

Suppose that for some $1 \leq i \leq n$, point $X_{i}$ lies on the segment $A_{i} P_{i}$, while point $X_{i+1}$ lies on the segment $P_{i+1} A_{i+2}$. Then the projection of the segment $X_{i} X_{i+1}$ onto the line $P_{i} P_{i+1}$ contains segment $P_{i} P_{i+1}$, since $\gamma_{i}$ and $\beta_{i+1}$ are acute angles (see Fig. 3). Therefore, $X_{i} X_{i+1} \cos \alpha_{i} \geq P_{i} P_{i+1}$, and in this case the statement is proved.

So, the only case left is when point $X_{i}$ lies on segment $P_{i} A_{i+1}$ for all $1 \leq i \leq n$ (the case when each $X_{i}$ lies on segment $A_{i} P_{i}$ is completely analogous).

Now, assume to the contrary that the inequality
$$\begin{equation*} X_{i} X_{i+1} \cos \alpha_{i}<P_{i} P_{i+1} \tag{1} \end{equation*}$$
holds for every $1 \leq i \leq n$. Let $Y_{i}$ and $Y_{i+1}^{\prime}$ be the projections of $X_{i}$ and $X_{i+1}$ onto $P_{i} P_{i+1}$. Then inequality (1) means exactly that $Y_{i} Y_{i+1}^{\prime}<P_{i} P_{i+1}$, or $P_{i} Y_{i}>P_{i+1} Y_{i+1}^{\prime}$ (again since $\gamma_{i}$ and $\beta_{i+1}$ are acute; see Fig. 4). Hence, we have
$$X_{i} P_{i} \cos \gamma_{i}>X_{i+1} P_{i+1} \cos \beta_{i+1}, \quad 1 \leq i \leq n .$$
Multiplying these inequalities, we get
$$\begin{equation*} \cos \gamma_{1} \cos \gamma_{2} \cdots \cos \gamma_{n}>\cos \beta_{1} \cos \beta_{2} \cdots \cos \beta_{n} . \tag{2} \end{equation*}$$
On the other hand, the sines theorem applied to triangle $P P_{i} P_{i+1}$ provides
$$\frac{P P_{i}}{P P_{i+1}}=\frac{\sin \left(\frac{\pi}{2}-\beta_{i+1}\right)}{\sin \left(\frac{\pi}{2}-\gamma_{i}\right)}=\frac{\cos \beta_{i+1}}{\cos \gamma_{i}} .$$
Multiplying these equalities we get
$$1=\frac{\cos \beta_{2}}{\cos \gamma_{1}} \cdot \frac{\cos \beta_{3}}{\cos \gamma_{2}} \cdots \frac{\cos \beta_{1}}{\cos \gamma_{n}}$$
which contradicts (2).

Fig. 3

Fig. 4