Is it true that for Normaliser of an element(defined below) to be a normal subgroup, it must be equal to the whole group? Normaliser of an element $a$ of a group G is defined as the set of all elements of G which commute with $a$.
$$N(a)=\{x\in G : xa=ax\}$$
As defined, it is easy to note that $N(a) < G. $ 
For $N(a)$ to be normal subgroup it should be true that, $$gxg^{-1} \in N(a)\textrm{ } \forall g \in G \textrm{ }\forall x \in N(a)$$ i.e.$gxg^{-1}a=agxg^{-1} \textrm{ }\forall g \in G \textrm{ }  \forall x \in N(a)$ Can I conclude from here that for this to be true, $g$ must commute with $a \textrm{ } \forall g\in G$ i.e. $N(a)=G$.

No. Take for example group of quaternions $G = \{1,-1, i,-i,  j,-j,  k, -k\}$. Then $N(i) = \{-1, 1, i, -i\}$. $N(i)$ is normal subgroup, since every subgroup of quaternions is.
Edit: this example shows flaw in your reasoning. You assumed, that if $A<B$, and $A$ is normal, then $\forall a\in A\forall \ b\in B:\ b^{-1}ab=a$, while it is only true, that $\forall a\in A \forall\ b\in B:\ b^{-1}ab\in A$. This is the case for quaternions: $$(-j)ij=-i\neq i$$
