My approach so far has been to separate $\sin(t)$ as a sum of exponentials to re-write the integral as a sum of two integrals where Jordan's Lemma can be easily applied for $p>0$:
$$\begin{align}\sin(t)&=\frac{1}{2i}(e^{it} - e^{-it})\\ \therefore \frac{sin(t)}{(t)}e^{ipt} &= \frac{e^{ipt}}{2ti}(e^{it} - e^{-it})\\ &= \frac{ie^{-it}}{2t}e^{ipt} - \frac{ie^{it}}{2t}e^{ipt}\end{align}$$
Hence, the original integral is re-writed as:
$$I = \int_{-\infty}^{\infty} \frac{\sin (t)}{t}e^{ipt} dt = \int_{-\infty}^{\infty} \frac{ie^{-it}}{2t}e^{ipt} dt - \int_{-\infty}^{\infty} \frac{ie^{it}}{2t}e^{ipt}dt$$
The issue I'm facing is trying to transform $f(t)$ into a complex valued function $f(z)$ to find the relevant singularities in order to compute their respective residues. My attempt (for the second integral) goes as follows:
$$z = e^{it}\therefore dz = ie^{it}dt\therefore dt = \frac{dz}{iz}$$
Additionally:
$$z = e^{it}\therefore it = \ln(z)\therefore t = -i\ln(z)$$
Substituting this results in $f(t)$:
$$\int_{-\infty}^{\infty} \frac{ie^{it}}{2t}e^{ipt}dt = \int_{-\infty}^{\infty} \frac{1}{-2i\ln(z)}e^{ipt}dz$$
But if this transformation is correct, I would face the issue of attempting to find the singularities of a multivalued function (the logarithm of a complex number) and I do not know as of now how to find them.
From this step, you can use Jordan's lemma immediately. For the second term, since $p>0$, you can choose the semi-circle on the upper half plane to apply Jordan's lemma. For the first term, you need to be careful,
(1) if $p=1$, the integrand is odd, hence the principle value is zero.
(2) if $p>1$, you choose the semi-circle on the upper half plane to apply Jordan's lemma.
(3) if $0<p<1$, you choose the semi-circle on the lower half plane to apply Jordan's lemma.