The circle containing semicircle $q$ is the image of $p$ in homothety with center $M$ and factor $1/2$, hence $Q$ is the midpoint of $RM$. Since triangles $MP_1Q$, $P_2RQ$ share the angle by $Q$, we have
$$\frac{S_1}{S_2} = \frac{\frac{1}{2} P_1 Q \cdot M Q \cdot \sin \angle P_1 Q M}{\frac{1}{2} P_2 Q \cdot R Q \cdot \sin \angle P_2 Q R} = \frac{P_1 Q}{P_2 Q}.$$
Let $KM = r$, $ML = x$, $P_1L = d_1$, $QL = d_2$ (Fig. 1). Points $P_1$, $P_2$ are symmetric about $KM$, therefore $P_1L = P_2L$ and $P_2Q = d_1 - d_2$. Denote by $M'$ the point such that $MM'$ is the diameter of $p$. Then triangle $M'MP_1$ is right and by Geometric Mean Theorem (an altitude splits a right triangle into two similar triangles) we get $d_1^2 = x(2r - x)$. Similarly in right triangle $KQM$ we get $d_2^2 = x(r - x)$ and thus
$$\begin{aligned} \frac{S_1}{S_2} &= \frac{P_1 Q}{P_2 Q} = \frac{d_1 + d_2}{d_1 - d_2} = \frac{(d_1 + d_2)^2}{d_1^2 - d_2^2} \\ &= \frac{x(2r - x) + x(r - x) + 2\sqrt{x(2r - x) \cdot x(r - x)}}{rx} \\ &= \frac{3r - 2x + 2\sqrt{(2r - x)(r - x)}}{r}. \end{aligned}$$

Fig. 1
We view the expression as a function of variable $x$ with parameter $r$. The function is decreasing on $(0, r)$ (both functions $3r - 2x$ and $(2r - x)(r - x)$ are decreasing), therefore it attains its maximum $3 + 2\sqrt{2}$ at $x = 0$ and minimum $1$ at $x = r$. By the problem statement, $x \in (0, r)$, thus $1 < S_1/S_2 < 3 + 2\sqrt{2} = 3 + \sqrt{8}$.