The tetrahedron has two equilateral faces whose sides are of length $2$, and two isosceles faces with two sides of length $2$ and one of length $1$. Let $F$ be the area of each equilateral face and $G$ the area of each isosceles face. It is obvious that $F > G$ holds. Further, let $a$ and $b$ be the distances of $P$ from the equilateral faces and $c$ and $d$ the distances from the isosceles faces.

If $V$ is the volume of the tetrahedron, we have
$$\begin{align*} 3V &= F(a+b) + G(c+d) = F(a+b+c+d) - (F-G)(c+d) \\ &\iff a+b+c+d = \frac{3V + (F-G)(c+d)}{F}. \end{align*}$$
Since $F - G > 0$, the value of $a+b+c+d$ is minimal for $c+d=0$, which is the case for $c=d=0$. The minimum value is therefore assumed for points $P$ on the common edge of the isosceles faces, i.e. on the edge with length $1$.

On the other hand, we also have
$$\begin{align*} 3V &= F(a+b) + G(c+d) = G(a+b+c+d) + (F-G)(a+b) \\ &\iff a+b+c+d = \frac{3V - (F-G)(a+b)}{G}. \end{align*}$$
Since $F - G > 0$, the value of $a+b+c+d$ is maximal for $a+b=0$, which is the case for $a=b=0$. The maximum value is therefore assumed for points $P$ on the common edge of the equilateral faces.