Is an expression in normal form if terminates on normal order but not applicative?
The expression $(\lambda x.y \  (\lambda x.(x \  \ x) \  \lambda x.(x \ \  x)))$
is not in normal form because it can be further reduced using the β-rule. 

Also, as you noted, Applicative Order is not a normalising reduction strategy. For the expression $(\lambda x.y \  (\lambda x.(x \  \ x) \  \lambda x.(x \ \  x)))$ it fails to find the normal form. This is because $(\lambda x.(x \  \ x) \  \lambda x.(x \ \  x))$ is a non-terminating argument, it has no normal form, then it will loop forever.

On the other side, Normal Order is a normalising reduction strategy because it will always find the normal form, if it exists of course. In fact this a practical consequence of the second Church-Rosser theorem. For the expression $(\lambda x.y \  (\lambda x.(x \  \ x) \  \lambda x.(x \ \  x)))$ it finds the normal form. In this case, since $\lambda x. y$ is a constant function, it will ignore its (non-terminating) argument.