Use:
$(i)$ the Ampere's law for $\vec{H}$ and
$(ii)$ continuity of lines of $\vec{B}$,to conclude that inside a bar magnet,
$(a)$ lines of $\vec{H}$ run from the $N$ pole to $S$ pole,while
$(b)$ lines of $\vec{B}$ must run from the $S$ pole to $N$ pole.

(N/A) Consider a magnetic field line of $\vec{B}$ forming a closed loop through the bar magnet as shown in the figure.
Let $C$ be an Amperian loop. According to Ampere's law for $\vec{H}$,the line integral of $\vec{H}$ over a closed loop is zero in the absence of free currents:
$\oint \vec{H} \cdot d\vec{l} = 0$
Inside the magnet,the relationship between $\vec{B}$ and $\vec{H}$ is given by $\vec{B} = \mu_0(\vec{H} + \vec{M})$.
Since magnetic field lines of $\vec{B}$ are continuous and form closed loops,they must travel from the $S$ pole to the $N$ pole inside the magnet to complete the loop outside from $N$ to $S$.
For the $\vec{H}$ field,the integral $\oint \vec{H} \cdot d\vec{l} = 0$ implies that if the path integral along the $\vec{B}$ line inside the magnet is positive,the path integral outside must be negative. Since $\vec{B} = \mu_0 \vec{H}$ outside the magnet (where $\vec{M}=0$),the direction of $\vec{H}$ lines inside the magnet must be from $N$ to $S$ to satisfy the closed loop condition for the total integral to be zero.