$\newcommand{\+}{^{\dagger}}
\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\down}{\downarrow}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\isdiv}{\,\left.\right\vert\,}
\newcommand{\ket}[1]{\left\vert #1\right\rangle}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
\newcommand{\wt}[1]{\widetilde{#1}}$
\begin{align}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{\rm P}\pars{\xi}&=
\int_{-\infty}^{\infty}\dd x\,{1 \over \root{2\pi}\sigma}\,
\exp\pars{-\,{x^{2} \over 2\sigma^{2}}}
\int_{-\infty}^{\infty}\dd y\,{1 \over \root{2\pi}\sigma}\,
\exp\pars{-\,{y^{2} \over 2\sigma^{2}}}\
\overbrace{\delta\pars{\xi - xy}}^{\ds{\delta\pars{y - \xi/x} \over \verts{x}}}
\\[3mm]&={1 \over 2\pi\sigma^{2}}\int_{-\infty}^{\infty}
\exp\pars{-\,{1 \over 2\sigma^{2}}\bracks{x^{2} + {\xi^{2} \over x^{2}}}}\,
{\dd x \over \verts{x}}
\\[3mm]&={1 \over \pi\sigma^{2}}\int_{0}^{\infty}
\exp\pars{-\,{1 \over 2\sigma^{2}}\bracks{x^{2} + {\xi^{2} \over x^{2}}}}\,
{\dd x \over x}\tag{1}
\end{align}
$\ds{\delta\pars{x}}$ is the
Dirac Delta Function.

With $\ds{x \equiv A\expo{\theta/2}\,,\quad A > 0\,,\quad\theta \in {\mathbb R}}$:
\begin{align}
{\rm P}\pars{\xi}&={1 \over \pi\sigma^{2}}
\int_{-\infty}^{\infty}
\exp\pars{-\,{1 \over 2\sigma^{2}}
\bracks{A^{2}\expo{\theta} + {\xi^{2} \over A^{2}}\expo{-\theta}}}\,
\pars{A\expo{\theta/2}\,\dd\theta/2 \over A\expo{\theta/2}}
\end{align}
We can choose $\ds{A}$ such that
$\ds{A^{2} = {\xi^{2} \over A^{2}}\quad\imp\quad A = \verts{\xi}^{1/2}}$:
\begin{align}
{\rm P}\pars{\xi}&={1 \over 2\pi\sigma^{2}}
\int_{-\infty}^{\infty}
\exp\pars{-\,{\verts{\xi} \over \sigma^{2}}\cosh\pars{\theta}}\,\dd\theta
={1 \over \pi\sigma^{2}}
\int_{0}^{\infty}
\exp\pars{-\,{\verts{\xi} \over \sigma^{2}}\cosh\pars{\theta}}\,\dd\theta
\end{align}

$$\color{#00f}{\large%
{\rm P}\pars{\xi} = {1 \over \pi\sigma^{2}}\,
{\rm K}_{0}\pars{\verts{\xi} \over \phantom{2}\sigma^{2}}}
$$
where $\ds{{\rm K}_{\nu}\pars{z}}$ is a
Second Kind Bessel Function.