Frequency conversion in IIR lowpass-to-lowpass transformation
The prototype filter is not uniquely defined by the two band edges of the filter to be designed. So you have to choose one of the two edge frequencies of the prototype filter. E.g., if you choose $\omega'_p$, you can compute the value of $\alpha$ according to
$$\alpha=\frac{\sin[(\omega'_p-\omega_p)/2]}{\sin[(\omega'_p+\omega_p)/2]}\tag{1}$$
where $\omega_p$ is the desired pass band edge. Now that you have $\alpha$, the stop band edge of the prototype filter can be computed from the desired stop band edge $\omega_s$:
$$\omega'_s=-\angle\left(\frac{e^{-j\omega_s}-\alpha}{1-\alpha e^{-j\omega_s}}\right)\tag{2}$$

EDIT: the lowpass-to-lowpass transformation
$$z^{-1}\longrightarrow\frac{z^{-1}-\alpha}{1-\alpha z^{-1}}\tag{3}$$
transforms a prototype edge frequency $\omega'_s$ in the following way:
$$e^{-j\omega'_s}=\frac{e^{-j\omega_s}-\alpha}{1-\alpha e^{-j\omega_s}}\tag{4}$$
From $(4)$ it follows that
$$-\omega'_s=\angle\left(\frac{e^{-j\omega_s}-\alpha}{1-\alpha e^{-j\omega_s}}\right)\tag{5}$$
which is equivalent to $(2)$.
Note that Eq. $(5)$ is different from Eq. $(8.71)$ in the book because $(8.71)$ is derived from a lowpass-to-highpass transformation.