Do I need to take elastic potential energy into account? 
Let's say I have a vertical spring with end points $A$ and $B$ and length $a$ and a mass
  attached to the endpoint $B$. The mass is dropped from the point $A$ and
  I need to find the kinetic energy of the mass at a point $ak$ where $k>1$.

Do I need to include the Elastic Potential Energy to find the kinetic energy? If yes, why can't I just use the Kinetic Energy and Gravitational Potential Energy?
I know $\text{Work Done}=mgka$ by the particle and the change in GPE is $mg(ka-a)$ but I don't get what else am I missing?
 A: It's up to you whether or not to include elastic potential energy. You'll get the same answer as long as you're careful about what you define to be inside and outside of your system.
The basic idea to use here is the work-energy theorem when only mechanical energy is of concern:
$$W_\text{net,external}=\Delta K_\text{total}+\Delta U_\text{mechanical}.$$
The subscript "$\text{external}$" is important on the left hand side; sum up all of the work done only by external forces (i.e., forces by external objects).


*

*If you choose the spring to be part of your system (so that it in considered internal), you would not include the work done by the spring in the $W_\text{net,external}$ term. In this case, you would interpret the spring as providing a change in potential energy $\Delta U_\text{elastic}$. One can only have a change in potential energy when the object(s) capable of "storing" potential energy are included in the system.

*If instead you choose the spring to be external to your system, then you would include the work done by the spring on your system in the $W_\text{net,external}$ term on the left-hand side. There would be no elastic potential energy change in your system since the spring is not part of your system.
That being said, most people would choose to include the block, spring, and Earth as the system. By making this choice, one can use the textbook expressions for elastic and gravitational potential energies. Additionally, there would be no external work done (since the other end of the spring doesn't move, presumably), so the left-hand side of the work-energy theorem is simpler with that choice.
Note that if you didn't include the Earth in your system, you would not have any gravitational potential energy change, but the Earth would do external work on your system.
