Solution

To solve this problem we can use the Lorentz transformation for momentum and energy in the horizontal direction.
The horizontal component of the momentum is:

\begin{equation} p_x = \frac{h \nu }{c} \cos\alpha \qquad E = h \nu \end{equation}

Everything else is the same as in problem $14.3.17$.

\begin{equation} h \nu' = \frac{h \nu - (-\beta c) p_x}{\sqrt{1-\beta^2}} = \frac{h \nu + h \nu \beta \cos\alpha}{\sqrt{1-\beta^2}} \rightarrow \nu' = \nu \frac{1+\beta \cos \alpha}{\sqrt{1-\beta^2}} \end{equation}

The negative sign appears because the velocity of the wall points in the negative direction.
The reflected wave will have this frequency; then transforming back to the Earth frame:

\begin{equation} h \nu_{\text{reflected}} = \frac{h \nu' + \beta c \frac{h \nu'}{c} \cos\alpha}{\sqrt{1-\beta^2}} \rightarrow \nu_{\text{reflected}} = \nu \frac{(1+\beta\cos\alpha)^2}{1-\beta^2} \end{equation}