How the braking radiation fit into the photon picture of light? The continuous  part of the  x ray spectrum is due the deceleration of electrons. I know that a decelerating charged particle emits a braking radiation according the EM theory. However, what's in the photon picture of light that says a decelerating charged particle must emit photons continuously? Why electrons colliding with the anode don't lose their energy exciting the atoms setting in there to some possible excited energy state instead?
 A: Classical models suffice for many cases of electromagnetic radiation. Bremsstrahlung is a phenomenon that's very difficult to approach from a quantum viewpoint, so we usually model it classically. Empirically, this is usually adequate. The classical theory predicts the intensity spectrum, and the intensity divided by the photon energy is the expected rate of detection of photons of that energy.

Why electrons colliding with the anode don't lose their energy exciting the atoms setting in there to some possible excited energy state instead?

They do. Most electrons colliding with the target in an x-ray tube don't lose much energy to bremsstrahlung. The majority of energy is lost to atomic excitations that cascade down into heat. Some atomic excitations radiate photons at characteristic x-ray energies, so there is generally a line spectrum added to the bremsstrahlung spectrum.
A: 
However, what's in the photon picture of light that says a decelerating charged particle must emit photons continuously?

I don't think the usual "photon picture" is compatible with "emitting continuously". The emission should be a series of well separable countable events to conform to the "photon picture". Emitting continuously evokes emission of a classical EM field wave that does not manifest any photon behaviour.
You may have meant "emit photons with high cadency". Then the photon picture would be as follows: from time to time, the electron bathing in external electric field and moving rapidly emits a photon; and energy and momentum for this newly created particle is taken from energy of the combined system (moving electron + EM field in its vicinity). In a sense, electron acts as a "condensation center" on which EM energy and EM momentum concentrates and gets moving as a particle. This EM energy and EM momentum come from EM energy and momentum in the space all around the electron in its vicinity. This energy and momentum is zero or unable to leave the electron when the electron is alone (free electron does not emit photons), but they are non-zero when the electron is in external field (this can be seen already in classical EM theory).
If instead of electron in static electric field, we have electron moving rapidly through a region filled with thermal EM radiation, then there can be a different kind of braking radiation: it is possible that the fast electron scatters a low energy photon into a high energy photon, and thus the electron loses kinetic energy while it produces "more visible" photons. This is called the inverse Compton effect.

Why electrons colliding with the anode don't lose their energy exciting the atoms setting in there to some possible excited energy state instead?

Electrons do not interact directly with atoms or other electrons, they interact with the immediate EM field where they are, and they interact with atoms only through this EM field. So naturally when electrons lose energy, this goes first into EM energy of radiation, and this spread in all directions, thus not all of it can be absorbed by the matter in the anode. Some of it may reach electrons in the atoms and produce characteristic radiation from there.
