Do photons lose energy after radiation pressure is applied to a perfect reflector?

In the case of a perfect reflector that does nothing but reflecting incoming radiation, it would seem the reflector is pushed and the radiation is reflected in opposite directions. But the radiation has not lost anything, it has only changed direction.

This is true only if the reflector has an infinitely high mass. Then the reflector doesn't begin to move.

In reality the reflector has a high but finite mass.

According to $p_{text{photon}}=frac{h}{lambda}$ you can set up the conservation of momentum $$frac{h}{lambda_{text{incident}}}=-frac{h}{lambda_{text{reflected}}}+mv$$ where $m$ is mass of the reflector, and $v$ is the velocity of the reflector after reflection. This means the reflector receives momentum from the incident photon. And thus the reflected photon has a momentum roughly the negative of the incident photon, but not exactly, as we see below.

And according to $E_{text{photon}}=hnu$ you can also set up the conservation of energy. $$hnu_{text{incident}}=hnu_{text{reflected}}+frac{1}{2}mv^2$$ This means the reflector also receives a tiny bit of energy from the incident photon. And thus the reflected photon has a tiny bit less energy than the incident photon. And thus its frequency $nu$ is a little less after reflection.

With the help $nu=frac{c}{lambda}$ this becomes $$frac{h}{lambda_{text{incident}}}=frac{h}{lambda_{text{reflected}}}+frac{mv^2}{2c}$$ and thus after reflection the photon's momentum is a little less and the photon's wavelength $lambda$ is a little longer.