Why is the K shell electron preferred in the photo electric effect? I have read in many books and on Internet as well that photoelectric effect is only possible when an electron is emitted from the K shell of the metal. Why not other bonded electrons?  
 A: As others have pointed out, the premise is false. However, there is still an element of truth to it, which is pretty easy to explain. It is true that when a photon has enough energy to ionize either a tightly bound electron or a weakly bound electron, it has a much higher probability of doing the former. This higher probability is shown by the fact that the K-shell edges are huge, constituing order-of-magnitude increases in the cross-section.
The reason for this is that you can estimate the cross-section using first-order perturbation theory, and it involves $|\langle i|eEz|f \rangle|^2$, where i is the electron's initial state (bound in an atom), and f is its final state (ionized). As discussed in more detail in the answers to this question, the transition matrix element tends to be small because the wavelength of a gamma or high-energy x-ray tends to be too small to be well matched with the spatial scale of the electron wavefunctions. You can say it either way around: the cross-section goes up with photon energy for a fixed electron orbital, or it goes down with electron energy for a fixed gamma energy.
A: The term "K-shell" stems from an older, now less used terminology for the 'electron shells' of multi-electronic atoms.
In this terminology, electrons with Principal Quantum Number $n$ equal to 1 where said to belong to the K-shell, those with $n=2$ the L-shell, those with $n=3$ the M-shell etc.
For an alkali metal like sodium, the electron configuration is $1s^22s^22p^63s^1$, so it has 2 electrons in the K-shell, 8 in the L-shell and 1 in the M-shell.
The inner electrons in the K and L-shells are much more tightly bound to the nucleus (due to electrostatic attraction between the positively charged nucleus and the negatively charged electrons) and cannot be 'knocked out' of their orbitals by visible light (which is not energetic enough). 
In the case of sodium only the unpaired $3s^1$ electron (M-shell) is energetically within reach of visible light photons because it is further away from the nucleus and has been shielded from electrostatic attraction by the K and L-shells.
This is generally true for all alkali metals, which have low ionisation energies due to the cited reasons. Alkali metals are therefore ideally suited to demonstrate the photo-electric effect.
A: The premise of your question is false. The photoelectric effect will occur whenever radiation interacts with the bounds electrons of an atom.
There is a sharp increase in absorption when the energy of the radiation exceeds the binding energy of a particular "shell". For Rb, for example, the NIST XCOM database shows clear K and L edges:

For atoms with lower Z (for example potassium) you don't always see such an edge except for the K shell electrons - it depends on the binding energies and available states. But the plot above shows it is possible to get photoelectric absorption from electrons in other shells.
A: In photo electric effect K-shell electron is favoured because of the high electron cloud density in the K-shell.
A: Most Photoelectric interactions occur in the K shell because the density of the electron cloud is greater in this region and a higher probability of interaction exists.  About 30% of photons absorbed from a dental x-ray beam are absorbed by the photoelectric process.
And secondly energy momentum conservation hold good for k-shell electron then other shells.
That's why 80% emissions occur from k-shell.
