Can protons have orbital if they gain more energy? I read that because a proton is much more massive than electron but an electron has slightly more energy so it doesn't fall into the nucleus and orbital is due to the constructive interference of electrons probability density cloud. Okay, but what about protons? Don't they also interference among each other constructively? Can I give each proton more energy so they can form tiny orbitals, and would this effects the electron orbitals in any way?
 A: Another view, let us take the hydrogen atom to avoid nuclear effects, and ask whether the proton can have an orbital.
What is an orbital? It is the quantum mechanical solution of the problem of the bound system of a proton and an electron which gives the probability  for an electron to be found at (x,y,z) given the system where the proton is at rest. This is a reasonable system since the mass of the proton is so much larger than the mass of the electron. If one took as (0,0,0) the center of mass system e+proton, then orbitals could be described for both the proton and the electron, except the proton orbitals would be confined in a very small volume, because of its large  mass.
It is analogous to the sun and planets, does the sun have an orbit? When the center of mass of the planetary system is taken into account, the center of mass of the sun has an orbit, but because of its mass, that orbit is very close or within the volume of the sun.
A: Protons absolutely can be excited into higher states, although we tend to think of these as excited states of the nucleus, rather than the atom.  The reasons is that in most cases, the spatial extent of the proton (or neutron; neutrons in the nucleus can also be excited into higher states) wave function is still tiny compared to the sizes of the electron wave functions (which set the size of the atom as a whole).  While the typical size of an atomic orbital is $10^{-10}\,{\rm m}$, the typical sizes of nuclear wave functions, even highly excited ones are $\lesssim 10^{-14}\,{\rm m}$.  So, regardless of whether the nucleus is in an excited state, the electrons still see the nucleus as an almost pointlike charge.
However, that "almost" is sometimes significant.  There are exceptions to what I described in the previous paragraph—such as for large-$Z$ atoms, for which the innermost $1S$ electrons are so tightly bound that they spend a significant fraction of their time inside the nucleus, which can lead to interesting mixing between atomic and nuclear excited states.  And even for isotopes of light elements like hydrogen and helium, it is possible to measure the effect of the finite size of the nucleus on the energies of the atomic $S$ states.  (This actually gives one of the most precise ways of measuring the radius of the proton.)  It is also possible to construct exotic artificial atoms, with the electron replaced by a much heavier muon, whose wave function is closer to the nucleus by a factor of $m_{\mu}/m_{e}\sim 200$—thus making the wave function $\sim(200)^{3}\sim10^{7}$ denser at the location of the nucleus and thus much more sensitive to nuclear size effects.
A: Within the nucleus, the protons and neutrons do strive to form shells with different energy levels, whereby they can all stack themselves together in a way that minimizes the binding energy cost. This is called the nuclear shell model.
