How P(C2H5)3 acts as a ligand? How does $\ce{P(C2H5)3}$ acts as a ligand and forms dπ-dπ bonds with transition elements. Why not pπ-dπ bond?  

I understand that transition elements have d subshell available to accept electrons. But what about the electrons from phosphorus? From Chemistry Class 12 (NCERT), Part 1, p. 168.:


  Another factor which affects the chemistry of nitrogen is the absence of d orbitals in its valence shell. Besides restricting its covalency to four, nitrogen cannot form dπ-pπ bond as the heavier elements can e.g., $\ce{R3P=O}$ or $\ce{R3P=CH2}$ ($\ce{R}$ = alkyl group). Phosphorous and arsenic can form dπ-dπ bond also with transition metals when their compounds like $\ce{P(C2H5)3}$ and $\ce{As(C6H5)3}$ act as ligands.
As I already mentioned in my comment, phosphines such as $\ce{PEt3}$ can bind to transition metals with partially filled d orbitals as a σ-donor as well as a π-acceptor.

The σ interaction is simply a donation of electrons from the lone pair on phosphorus to empty metal d orbitals.



The π interaction is backdonation of electrons from filled metal d orbitals to an empty orbital on the phosphine. Traditionally this was thought to be the 3d orbitals on phosphorus, hence "dπ–dπ". However, as Martin pointed out in the comments this is a rather outdated mode of thinking. The current accepted view is that the acceptor orbitals are $\ce{P-R}$ σ* orbitals (to be precise, a linear combination of them that has appropriate symmetry to overlap with metal d orbitals).