According to question, the given situation is shown in the figure below.

For a sufficiently safe-horizontal displacement,  $∆$s can be considered straight. If the corresponding length of path element is $∆$L, the work done against frictional force is given by

$w_{∆F}=\mu mg\left(\Delta L\cos \theta \right)$

From   $\Delta OPQ,\cos \theta =\frac{\Delta s}{\Delta L}$

So,   $W_{∆F}=\mu mg\Delta L\left(\frac{\Delta s}{\Delta L}\right)=\mu mg\Delta s$

On adding up, we find that along the whole path, the total work done by the friction force is $\mu$mgs. By law of conservation of energy, this must equals to the decrease in potential energy of skier.

$\therefore \mu mgs=mgh$

Hence,    $h=\mu s$