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Without loss of generality, we can create spherical coordinates such that $\mathbf{p}$ is along the $\theta=0$ axis. Then, we can transform the integral into those coordinates to obtain \begin{align} D(x-y) &= \frac{1}{(2\pi)^3}\, \int_0^\infty \int_0^\pi \int_0^{2\pi} \frac{1}{2E_{\mathbf{p}}} e^{i\, p\, r\, \cos\theta}\, r^2\, \sin\theta\, d\phi\, d\theta\, dr \\ &= \frac{1}{(2\pi)^2}\, \int_0^\infty \frac{r^2}{2E_{\mathbf{p}}} \left[\int_0^\pi e^{i\, p\, r\, \cos\theta}\, \sin\theta\, d\theta\right]\, dr. \end{align} Now, the integral in brackets is not hard, but is left as an exercise for the reader, who might note that $\frac{ e^{i p r}- e^{-i pr}}{2ipr} = \frac{ \cos(pr)}{ipr}$. Finally, swap $p$ and $r$, and you get the desired result. Of course, that last step might not work of $E_{\mathbf{p}}$ depends on some combination of $\mathbf{p}$ and $\mathbf{r}$ other than $pr$...