√(M1+M2+...+M100)

Draw a circle with radius $2$ and center $A$ . Then $\overline{BC}$ is a chord of the circle, and for any $i$ the segment $\overline{AP_i}$ extends to a diameter of the circle $\overline{S_iAP_iT_i}.$ It follows from the power of a point theorem for two secants that $S_iP_i \cdot P_iT_i= BP_i\cdot P_iC$ Therefore, $4 - AP_i^2 = (2+AP_i)\cdot (2-AP_i) = S_iP_i \cdot P_iT_i= BP_i\cdot P_iC \implies M_i = AP_i^2 + BP_i \cdot P_iC = 4$

Since this is true for all $i$ , $\sqrt{M_1 + M_2 + \cdots + M_{100}} = \sqrt{100 \cdot 4} = \boxed{20}$

$\large\begin{aligned}\hspace{20mm}\text{Consider }&\Delta ABC\\\\ AB&=AC=2\hspace{4mm}\color{#3D99F6}\small\text{ (Given) }\\ M_{i}&=(AP_{i})^2+BP_{i}\times CP_{i}\\\\ \text{Construct } AD&\perp BC\\ \text{We have, }\\ BD&=CD\hspace{4mm}\color{#3D99F6}\small\text{Since }\Delta ABC \text{ is isoceles. }\\\\ \text{Let, } \\ BD&=CD=x\\ P_{i}D&=x_{i}\\ M_{i}&=(AP_{i})^2+BP_{i}\times CP_{i}\\ &=(AP_{i})^2+(x-x_{i})\times (x+x_{i})\\ &=(AP_{i})^2+(x^2-{x_{i}}^2)\\ &=AD^2+x^2\hspace{4mm}\color{#3D99F6} \small \text{Since }AP_{i}^2-{x_{i}}^2=AP_{i}^2-{P_{i}D}^2=AD^2\\ &=AB^2=4\\ \implies M_{i}&=4\\ \implies \sqrt{\sum_{i=1}^{100}M_{i}}&= \sqrt{4 \times100}=\color{#EC7300}\boxed{\color{#333333}20} \end{aligned}$