This argument involving $\Gamma(\omega)$ is made since in a naive approach, in order to obtain the Hawking radiation, one generally drops the effective potential. Let me make this statement clearer:
In a curved spacetime, the Lagrangian of a free massless scalar field is:
\begin{equation}
\mathcal{L}=\frac{1}{2}g^{\mu\nu}
\partial_\mu\phi\,\partial_\nu \phi
\end{equation}
where the metric $g^{\mu\nu}$ is the solution of the Einstein equations:
$$G_{\mu\nu}=8\pi G\,T_{\mu\nu}$$
in a spacetime with a source $T_{\mu\nu}$.
Now, by making use of the Euler-Lagrange equations, you can write down the equations of motion, which reads:
$$\hat{\square}\phi=0$$
where $\hat{\square}$ is the Laplace-Beltrami operator:
$$\frac{1}{\sqrt{g}}\partial_\mu (\sqrt{g}\,g^{\mu\nu}\partial_\nu)$$
with $g=det(g_{\mu\nu})$.
If you now consider the Schwarzchild metric given by
$$ds^2=(1-\frac{2m}{r})dt^2-\frac{dr^2}{1-\frac{2m}{r}}-r^2d\Omega^2$$
Since we are in a spherically symmetric spacetime, we can expand the field in spherical harmonics:
$$\phi=\sum_{l,m}\frac{F(r,t)}{r}Y_l^m(\theta,\phi)$$
And now, by putting this expansion in the equation of motion, and considering the radial part, you will obtain (after some algebraic steps):
$$\left(\frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial {r^*}^2}+V_l\right)F_l(r,t)=0$$
with the effective potential $V_l=(1-\frac{2m}{r})(\frac{2m}{r^3}+\frac{l(l+1)}{r^2})$.
In a first approximation, you can drop the $V_l$ term, by saying that your solution will be valid in the $r\to\infty$ regions (the asympotical regions). What you will find is a mean particle value given by the planckian distribution at late time:
$$\langle in|N^R_\omega|in\rangle=\frac{1}{e^{8\pi m\omega}-1}$$
(where $N^R_\omega={a^R_\omega}^\dagger a^R_\omega$, with this creation and annhilation operators referring to the modes $u^R_\omega\sim e^{-i\omega u}$ and with $u=t-r^*$, one of the null coordinates in the double-null extension of the Schwarzchild metric).
This value is clearly divergent for $\omega\to 0$, and this is clear since we have dropped the effective potential, which would shield the modes allowed to go to the infinity. If you consider the effective potential, the equation of motion will obviously change, and you can say that there will be some Transmitted modes (T-modes) and some Reflected modes (R-modes) by the potential. Since we are considering modes $\sim e^{-i\omega u}$, which are directed through the future null infinity $\mathcal{I^+}$ (in a penrose diagram), only the T-modes will arrive at this $\mathcal{I}^+$, since the R-modes will be reflected by the potential into the Black-Hole. Thus, by an asymptotic analysis, you can calculate this $T$ and $R$ probability, and you will find that:
$$|T|^2\simeq 16m^2\omega^2\sim A_H\omega^2\;\;\;\; \text{with }A_H\text{ the horizon surface area}$$
(and $|R^2|=1-|T^2|$).
Your Planckian distribution will then become:
$$\langle in|N^R_\omega|in\rangle=\frac{|T|^2}{e^{8\pi m\omega}-1}=\frac{\Gamma(\omega)}{e^{8\pi m\omega}-1}$$
and this is correct since the low modes are shielded by this $\Gamma$.
However, in your reference, Parker is considering the modes $\sim e^{-i\omega v}$, with $v=t+r^*$, the other null-coordinate of the double-null extension of the Schwarzchild metric, then you have that the modes which will go to infinity, will be those that are reflected by the potential, and by doing the same analysis I have done, you will arrive at the same conclusion.
