These kinds of logs appear whenever you have things like $a^\epsilon$ where $\epsilon$ is a small parameter we are expanding in powers of. In fact, by definition $a^\epsilon = \exp(\epsilon \log a)$ and therefore we have the expansion $$a^\epsilon = 1+\epsilon\log a + O(\epsilon^2)\tag{1}.$$
That is exactly the kind of thing that is going on here. When doing dimension regularization we write, using PS conventions, $d = 4-\epsilon$ and take the Feynman diagram, which is a function of $d$, and expand in powers of $\epsilon$. For the precise term you are considering, we see that: \begin{eqnarray}\Gamma\left(2-\frac d2\right)\left(\frac 1 \Delta\right)^{2-\frac d2}&=&\Gamma\left(\frac{\epsilon}{2}\right)\Delta^{-\frac{\epsilon}{2}}\\ &=&\left(\frac{2}{\epsilon}-\gamma+O(\epsilon)\right)\left(1-\frac{\epsilon}{2} \log \Delta+O(\epsilon^2)\right)\\ &=&\frac{2}{\epsilon}-\log\Delta-\gamma+O(\epsilon)\tag{2}.\end{eqnarray}
That explains to you the $\log \Delta$ part of the question. The $\log(4\pi)$ does not come from here. Its origin is the factor of $\frac{1}{(4\pi)^{d/2}}$ that appears when evaluating the Feynman diagrams after Wick rotation. See for example the unnumbered equation in page 250 of PS below the definition of the Euler beta function: $$\int \dfrac{d^d\ell_E}{(2\pi)^d}\dfrac{1}{(\ell_E^2+\Delta)^2}=\dfrac{1}{(4\pi)^{d/2}}\dfrac{\Gamma(2-\frac{d}{2})}{\Gamma(2)}\left(\frac{1}{\Delta}\right)^{2-\frac{d}{2}}.\tag{3}$$
As you can see, compared to what you were considering there is that prefactor of $\frac{1}{(4\pi)^{d/2}}$. It can be expanded using (1): $$\frac{1}{(4\pi)^{d/2}}=\frac{1}{(4\pi)^{2-\epsilon/2}}=\frac{1}{(4\pi)^2}(4\pi)^{\epsilon/2}=\frac{1}{(4\pi)^2}\left(1+\frac{\epsilon}{2}\log (4\pi)+O(\epsilon^2)\right)\tag{4}.$$
Multiplying (2) and (4) together gives you equation (7.84) in PS:
$$\int \dfrac{d^d\ell_E}{(2\pi)^d}\dfrac{1}{(\ell_E^2+\Delta)^2}=\dfrac{2}{\epsilon}-\log \Delta-\gamma+\log(4\pi)+O(\epsilon)\tag{5}$$
