As noted by @copper.hat in the comments above, your main problem is that you're using the variable $\omega$ twice. The dummy variable of integration in your very first expression should be given a different name, $\omega'$ say:
\begin{equation}
x_p(t) \;=\; \int_0^{\infty}d\omega'\, D(\omega')\, \cos[\omega' t - \phi(\omega')]
\end{equation}
Then:
\begin{align*}
FT\left\{x_p(t)\right\}
&=
\frac{1}{2\pi}\int_{-\infty}^{+\infty} dt\, e^{-i \omega t}\, x_p(t)\\[0.1in]
&=
\frac{1}{2\pi}\int_{-\infty}^{+\infty} dt\, e^{-i \omega t}
\int_0^{\infty}d\omega'\,D(\omega')\, \cos[\omega' t - \phi(\omega')]\\[0.1in]
&=
\int_0^{\infty}d\omega'\,D(\omega')\,\frac{1}{2\pi}\int_{-\infty}^{+\infty} dt\, e^{-i \omega t}\cos[\omega' t - \phi(\omega')]\\[0.1in]
&=
\int_0^{\infty}d\omega'\,D(\omega')\,\frac{1}{2}\,\frac{1}{2\pi}\int_{-\infty}^{+\infty} dt\, e^{-i \omega t}
\left[e^{+i\omega't}e^{-i \phi(\omega')} \,+\,e^{-i\omega't}e^{+i \phi(\omega')}\right]\\[0.1in]
&=
\frac{1}{2}\int_0^{\infty}d\omega'\,D(\omega')\,\left[\delta(\omega' - \omega)e^{-i \phi(\omega')} \,+\,\delta(\omega'+\omega)e^{+i \phi(\omega')}\right]\\[0.1in]
&= \frac{1}{2}\times
\begin{cases}
D(\omega)\, e^{-i \phi(\omega)} & \omega > 0\\[0.05in]
D(-\omega)\, e^{+i \phi(-\omega)} & \omega < 0\\[0.05in]
D(0)\,\cos\phi(0) & \omega = 0
\end{cases}
\end{align*}
In going from the third to the fourth line above, I have used the complex expression for $\cos$:
$$
\cos x = \frac{e^{+i x} - e^{-i x}}{2}\, .
$$
In going from the fourth to the fifth line above, I have used the integral representation of the Dirac delta function:
$$
\delta (\omega) = \frac{1}{2\pi} \int_{-\infty}^{+\infty}dt\, e^{\pm i \omega t}
$$
