= Solution
With the specified <Fourier transform> convention, the inverse is $\delta(\mathbf x)=\int d^3k\,\delta_{\mathbf k}e^{-i\mathbf k\cdot\mathbf x}/(2\pi)^3$. Inserting both inverse transforms into the <cosmological two-point correlation function> and applying the given <Dirac delta> covariance yields
$$
\begin{aligned}
\xi(\mathbf r)
&=\int\frac{d^3k\,d^3k'}{(2\pi)^6}
e^{-i\mathbf k\cdot\mathbf x-i\mathbf k'\cdot(\mathbf x+\mathbf r)}
\langle\delta_{\mathbf k}\delta_{\mathbf k'}\rangle\\
&=\int\frac{d^3k}{(2\pi)^3}P(k)e^{i\mathbf k\cdot\mathbf r}.
\end{aligned}
$$
For an isotropic <matter power spectrum>, align the polar axis with $\mathbf r$. The angular integral is $2\pi\int_{-1}^1e^{ikr\mu}\,d\mu=4\pi\sin(kr)/(kr)$. Hence the <isotropic cosmological correlation-power-spectrum relation> is
$$
\boxed{\xi(r)=\frac1{2\pi^2}\int_0^\infty k^2P(k)\frac{\sin(kr)}{kr}\,dk.}
$$
The angular kernel is the order-zero <Spherical Bessel function> $j_0(kr)$. At $r=0$, replace it by its <limit> one, giving the unsmoothed <variance> when the integral converges. The argument assumes an ordinary integrable spectrum, or an explicitly specified <distributional Fourier transform> interpretation otherwise.
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