> $K.E=\frac{1}{2}mv^2$, $P=mv$ 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.
$$\vec{P}=m_1\vec{v}_1+m_2\vec{v}_2 \tag{1}$$
It is this quantity which is conserved during the collision.

Likewise you need to consider the total kinetic energy of both bodies.
$$E_\text{kin}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$
It is this quantity which is conserved during an *elastic* collision.

So there are many possible ways how the velocities $\vec{v}_1$ and $\vec{v}_2$
can change to make the kinetic energy $E_\text{kin}$ decrease while still
preserving the total momentum $\vec{P}$.

> 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
$$\vec{F}_{2\to 1}=-\vec{F}_{1\to 2}$$
With Newton's second law ($\vec{F}=m\frac{\Delta \vec{v}}{\Delta t}$)
you further get
$$m_1\frac{\Delta \vec{v}_1}{\Delta t}=-m_2\frac{\Delta \vec{v}_2}{\Delta t}$$

Rearranging this you get
$$\frac{\Delta(m_1\vec{v}_1+m_2\vec{v}_2)}{\Delta t}=\vec{0}$$
and hence
$$m_1\vec{v}_1+m_2\vec{v}_2=\text{const}$$
which is just the above mentioned conservation of total momentum.