Let $H_C$ be the foot of $C$ on line $AB$. Let $\Omega$ be the nine point circle of triangle $ABC$. Of course it consists points $K,L,M,H_C$. Denote $L'$ as the intersection of $LX$ and $BC$. Circles $\Omega,\Gamma$ are externally tangent, so there exists homothety $\kappa$ centered at $X$ such that $\kappa(\Gamma)=\Omega$. Because points $L',X,L$ are collinear and $L\in\Omega,L'\in\Gamma$ we have $$\kappa(L')=L$$$BC||ML$ so segments $L'B,ML$ are parallel chords of circles $\Gamma,\Omega$ respectively. Thus $$\kappa(L'B)=ML$$Both reached equalities lead to $$\kappa(B)=M$$. Hence $B,X,M$ are collinear $$X\in AB$$. From the definition of the point $$X\in \Omega$$. Finally $$X=H_C$$.

Is it Externally tangent or internally tangent

Internally when $\angle ABC>90^\circ$ and externally when $\angle ABC<90^\circ$. When $\angle ABC=90^\circ$ the circle $\Gamma$ is a point $X=B$.

I think this case is internally because the nine point circle Is inside the Circle $\Gamma$

The problem doesn't need to mention whether it's internally or externally tangent case, because that is dependent only on the position of $B$ with respect to the nine point circle. My solution is correct for all cases and so is the problem.