= Solution
Use <Fourier inversion> with the stated <Fourier transform> convention:
$$
f(x)=\frac1{2\pi}\int_{-\infty}^{\infty}\frac{-2ik}{p^2+k^2}e^{ikx}\,dk.
$$
For $x<0$, close the <contour integral> in the lower half-plane, where $|e^{ikx}|=e^{-x\operatorname{Im}k}$ 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> has
$$
\operatorname{Res}_{k=-ip}\left(\frac{-2ik}{(k-ip)(k+ip)}e^{ikx}\right)
=-ie^{px}.
$$
Consequently
$$
\boxed{f(x)=-e^{px},\qquad x<0.}
$$
The minus sign uses both the clockwise orientation and the residue $-i e^{px}$; it is consistent with the full odd function $\operatorname{sgn}(x)e^{-p|x|}$ away from zero.
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