The original form of the integral does not converge due to a non-integrable singularity at $p=0$.
The other two forms assume that you're taking the Cauchy principal value of this integral, or, equivalently, taking the real part of the integral.
To evaluate the integral in its third form, all we need is the the Laplace transform of $\frac{\sin(bp)}{p(e^{cp}+1)} $, which can be expressed in terms of the log gamma function.
I'll show how to find it.
Assume that $b \in \mathbb{R}$, $c>0$, and $a +c >0$.
The Laplace transform of $\frac{\sin(bp)}{e^{cp}+1}$ is
$ \begin{align} \int_{0}^{\infty} \frac{e^{-ap}\sin(bp)}{e^{cp}+1} \, \mathrm dp &= \frac{1}{2i}\int_{0}^{\infty} \frac{e^{-(a-ib)p}-e^{-(a+ib)p}}{e^{cp}+1} \, \mathrm dp \\ &= \frac{1}{2i}\int_{0}^{\infty} \frac{e^{-(a-ib+c)p}-e^{-(a+ib+c)p}}{1+e^{-cp}} \, \mathrm dp \\ &= \frac{1}{2i}\int_{0}^{\infty} \left( e^{-(a-ib+c)p} - e^{-(a+ib+c)p}\right) \sum_{n=0}^{\infty} (-1)^{n}e^{-cnp} \, \mathrm dp \\ &= \frac{1}{2i}\sum_{n=0}^{\infty} \int_{0}^{\infty} \left( e^{-(a-ib+c+cn)p} -e^{-(a+ib+c+cn)p} \right)\\ &= \frac{1}{2i}\sum_{n=0}^{\infty}(-1)^{n} \left( \frac{1}{a-ib+c+cn} - \frac{1}{a+ib+c+cn}\right)\\ &= \frac{1}{2ic} \left( \sum_{n=0}^{\infty} \frac{(-1)^{n}}{\frac{a-ib}{c}+1+n} - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{\frac{a+ib}{c}+1+n}\right)\\ &\overset{(1)}{=} \small\frac{1}{4ic} \left(\psi \left(\frac{a-ib}{2c}+1 \right)- \psi \left(\frac{a-ib}{2c} + \frac{1}{2} \right) - \psi \left(\frac{a+ib}{2c}+1 \right) + \psi \left(\frac{a+ib}{2c} + \frac{1}{2} \right)\right). \end{align}$
Therefore, the Laplace transform of $\frac{\sin(bp)}{p(e^{cp}+1)} $ is $$ \begin{align} &\int_{0}^{\infty} \frac{e^{-ap}\sin(bp)}{p(e^{cp}+1)} \, \mathrm dp \\ &\overset{(2)}{=} \small\frac{1}{4ic}\int_{a}^{\infty} \left(\psi \left(\frac{t-ib}{2c}+1 \right)- \psi \left(\frac{t-ib}{2c} + \frac{1}{2} \right) - \psi \left(\frac{t+ib}{2c}+1 \right) + \psi \left(\frac{t+ib}{2c} + \frac{1}{2} \right)\right) \, \mathrm dt \\ &= \small \frac{1}{2i} \left( \log \Gamma \left(\frac{t-ib}{2c}+1 \right) - \log \Gamma \left(\frac{t-ib}{2c} \frac{1}{2} \right)- \log \Gamma \left(\frac{t+ib}{2c}+1 \right) + \log \Gamma \left(\frac{t+ib}{2c}+\frac{1}{2} \right)\right) \Bigg|^{\infty}_{a} \\ &\overset{(3)}{=} \small \frac{1}{2i} \left( -\log \Gamma \left(\frac{a-ib}{2c}+1 \right) + \log \Gamma \left(\frac{a-ib}{2c} + \frac{1}{2} \right)+\log \Gamma \left(\frac{a+ib}{2c}+1 \right) - \log \Gamma \left(\frac{a+ib}{2c} + \frac{1}{2} \right)\right) \\ &= \Im \left(\log \Gamma \left(\frac{a+ib}{2c} + 1 \right) - \log \Gamma \left(\frac{a+ib}{2c} + \frac{1}{2} \right)\right) . \end{align}$$
$(1)$ https://mathworld.wolfram.com/DigammaFunction.html (6)
$(2)$ https://en.wikipedia.org/wiki/Laplace_transform#Evaluating_improper_integrals
$(3)$ As $t \to +\infty$, $$\log \Gamma(t+ \alpha) - \log \Gamma(t+ \beta) = (\alpha - \beta) \log(t) + O \left(\frac{1}{t} \right).$$
Therefore, as $t \to +\infty$, $$\small \log \Gamma \left(\frac{t-ib}{2c}+1 \right) - \log \Gamma \left(\frac{t-ib}{2c} + \frac{1}{2} \right)-\log \Gamma \left(\frac{t+ib}{2c}+1 \right) + \log \Gamma \left(\frac{t+ib}{2c} + \frac{1}{2} \right) \sim O \left(\frac{1}{t} \right). $$
So for $c>0$, $0 < \theta < \pi/2$, and $2c - \sin (\phi) >0 $, we have
$$\int_0^{\infty} \frac{e^{-p \sin (\phi)} \sin (p \cos (\phi ))}{p \left(e^{c p}+1\right)} \, \mathrm dp = \Im \left(\log \Gamma \left(\frac{ie^{- i \phi}}{2c} +1 \right) - \log \Gamma \left(\frac{ie^{-i \phi}}{2c} + \frac{1}{2} \right)\right)$$ and
$$\int_0^{\infty} \frac{e^{-p (c - \sin (\phi))} \sin (p \cos (\phi ))}{p \left(e^{c p}+1\right)} \, \mathrm dp = \Im \left(\log \Gamma \left(\frac{ie^{i \phi}}{2c} +\frac{3}{2} \right) - \log \Gamma \left(\frac{ie^{i \phi}}{2c} + 1 \right)\right). $$
Best Answer
This calculation will use the definition of the Exponential Integral and its Puiseux Series:
$$\int_{ }^{ }\frac{e^{x}}{x}dx = \operatorname{Ei}(x) + C \text{ and } \operatorname{Ei}(z) = \gamma + \log(z) + \sum_{n=1}^{\infty}\frac{z^{n}}{nn!}.$$
Define $\displaystyle I = \int_{-\infty}^{\infty}\frac{\left(e^{c\left|x\right|}-1\right)e^{aix-b\left|x\right|}}{x}dx$ where $a=\cos(\phi)$ and $b = \sin{(\phi)}$. We can define the parameters for $\phi$ and $c$ as the ones you listed in the post above. Since the limit as $x \to 0$ of the integrand does not exist but the integrand is improperly integrable on $\mathbb{R}$, we can split up the integral to apply FTC safely like this:
$$\int_{-\infty}^{\infty}\frac{\left(e^{c\left|x\right|}-1\right)e^{aix-b\left|x\right|}}{x}dx=\int_{-\infty}^{0}\frac{\left(e^{c\left|x\right|}-1\right)e^{aix-b\left|x\right|}}{x}dx+\int_{0}^{\infty}\frac{\left(e^{c\left|x\right|}-1\right)e^{aix-b\left|x\right|}}{x}dx.$$
We can calculate the first integral and deal with the removable discontinuity at $x=0$ in this evaluation:
$$ \begin{align} I_1 :=& \int_{-\infty}^{0}\frac{\left(e^{-cx}-1\right)e^{aix+bx}}{x}dx \\ =& \Big[\operatorname{Ei}(aix+bx-cx)-\operatorname{Ei}(aix+bx)\Big]_{-\infty}^{0} \\ =& \Bigg[\gamma + \log((ai+b-c)x) + \sum_{n=1}^{\infty}\frac{(ai+b-c)^n x^n}{nn!} \\ &-\gamma - \log((ai+b)x) - \sum_{n=1}^{\infty}\frac{(ai+b)^n x^n}{nn!} \Bigg]_{-\infty}^{0} \\ =& \Bigg[\log\left(ai+b-c\right)+\sum_{n=1}^{\infty}\frac{\left(ai+b-c\right)^{n}x^{n}}{nn!}-\log\left(ai+b\right)-\sum_{n=1}^{\infty}\frac{\left(ai+b\right)^{n}x^{n}}{nn!}\Bigg]_{-\infty}^{0} \\ =& \log\left(ai+b-c\right)-\log\left(ai+b\right) \\ =& \ln\left|ai+b-c\right|+i\operatorname{arg}\left(ai+b-c\right)-\ln\left|ai+b\right|-i\operatorname{arg}\left(ai+b\right) \\ =& \ln\sqrt{a^{2}+\left(b-c\right)^{2}}+i\arctan\left(\frac{a}{b-c}\right)-\ln\sqrt{a^{2}+b^{2}}-i\arctan\left(\frac{a}{b}\right). \\ \end{align} $$
I can leave $I_2$ as an exercise, but if anyone wants to spoil themselves, it is
We can combine these two results to get
$$I = 2i\arctan\left(\frac{a}{b-c}\right)-2i\arctan\left(\frac{a}{b}\right).$$
We can conclude by multiplying $i$ on both sides to get the desired result, which is
$$\int_{-\infty}^{\infty}\mathrm{d}p \frac{i \left(e^{c | p| }-1\right) e^{-| p| \sin (\phi )+i p \cos (\phi )}}{p} = 2\arctan\left(\cot\phi\right)-2\arctan\left(\frac{\cos\phi}{\sin\phi-c}\right).$$