> $K.E=\frac{1}{2}mv^2$, $P=mv$ 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.