$K.E=\frac{1}{2}mv^2$, $P=mv$ thus there is a relation between them. I really cant understand why there is no decrease in momentum when kinetic energy is decreased in inelastic collision.
It doesn't make sense to consider a single body during a collision (i.e. during interaction with a second body). You need to look at both bodies together. Therefore you need to consider the total momentum of both bodies. $$P=m_1v_1+m_2v_2 \tag{1}$$ It is this quantity which is conserved during the collision.
Likewise you need to consider the total kinetic energy of both bodies. $$K.E.=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$ It is this quantity which is conserved during an elastic collision.
Should i leave this question until i learn "Lagrangian" and "Noether's Theorem" ?
To understand momentum conservation (1) you don't need Lagrangian mechanics or Noether's theorem. Newton's mechanics is just enough.
According to Newton's third law (actio = reactio) you have $$F_{2\to 1}=-F_{1\to 2}$$ With Newton's second law ($F=m\frac{\Delta v}{\Delta t}$) you further get $$m_1\frac{\Delta v_1}{\Delta t}=-m_2\frac{\Delta v_2}{\Delta t}$$
Rearranging this you get $$\frac{\Delta(m_1v_1+m_2v_2)}{\Delta t}$$ and hence $$m_1v_1+m_2v_2=\text{const}$$ which is just the above mentioned conservation of total momentum.