 Solution

Use Fourier inversion with the stated Fourier transform convention:For $x<0$, close the contour integral in the lower half-plane, where $|e^{ikx}|=e^{-x\mathrm{Im}k}$ decays. Jordan lemma eliminates the large semicircle. The contour is clockwise, so the residue theorem supplies $-2\pi i$ times the residue at $k=-ip$. That simple pole hasConsequentlyThe minus sign uses both the clockwise orientation and the residue $-i{e}^{px}$; it is consistent with the full odd function $\mathrm{sgn}(x)e^{-p|x|}$ away from zero.