Net work done on the body when we lift it and put it on the table is zero? I'm little confused here. Work done on the body when we lift it and put it on the table is zero, because according to work energy theorem, change in kinetic energy of the body is zero. So, the net work done is zero. 
Fine, but now the object has $mgh$ (where $m=$mass of body; $h=$height of body) amount of potential energy stored in it. If the net work done is zero then who has increased its energy?
I'm confused and unable to think it through kindly help.
 A: You did net work on the body.  Gravity did negative net work on the body.  The over all work done was zero.  The original confusion arose because the work-energy theorem demands we calculate the change in kinetic energy using the net force on the body, but your question considered only the force exerted by you, and ignored that exerted by gravity.
If  the body has mass $m$, you were putting a force $mg$ on it to raise it at constant speed.  The work-energy theorem says that if you had done this when there were no other forces on the body, the body would have gained kinetic energy $mgh$ as you moved it from the floor to the table.
That analysis ignores gravity, though.  Gravity pulled down on the box with force $-mg$.  This means the work done by gravity was $-mgh$, and so the total work done on the box was zero.  This makes sense because if you lift the box at constant speed, the net force on the box is zero by $F = ma$.  
If the box starts by sitting stationary on the floor, there would have to be some small net work done on the box to get it started going up.  There would  have to be some small negative net work on it to get it to stop.
A: To answer such a question one must define the system under consideration.
System - body alone
An upward force $F$ lifts the body by a vertical distance $h$.
The two external forces acting on the body are the force $F$ and the weight of the rock $W=mg$.
The net work done on the body is $(F-W)h$ and this is equal to the change in the kinetic energy of the body which is zero if the two external forces are equal in magnitude.
System - body and Earth
The only external force is $F$ and let this force have the same magnitude as the weight of the body.
That single force $F$ cannot move the body further away from the Earth.
All that force can do is move the centre of mass of the body and Earth system.
To separate the body and the Earth two forces of magnitude $F$ but opposite in direction must act on the body and the Earth.
Because the mass of the Earth is so much greater than the mass of the body the displacement of the Earth relative to the centre of mass of the Earth and body system is very small compared to the displacement of the body.
The work done on the Earth by the external forces is negligible compared with the work done on the body.
The work done by the external forces on the Earth and body system is $Fh$.
The increase in gravitational potential energy of the system is $Fh$ which is $mgh$.  
What is often not appreciated is that it is the body and the Earth system which has the gravitational potential energy not the body alone.
As a result of the mass of the Earth being so much larger than the mass of the body the motion of the Earth is always so much smaller than the motion of the body and so the motion (change in kinetic energy) of the Earth is ignored (assumed to be negligible) compared with the motion (change in kinetic energy) of the body.
A: Don't complicate things for youself.Net work done is a simple term , which means the transfer in energy to do certain work(where energy transfers , not transform into its original form) .In the above case energy is transferred and then transform into its original form (another e.g is "SPRING"). 
A: There is work because the box transfer from one place to another which means there is a change in position.   
