Solution:

Consider fixing $a$ and $b$. Then, we know that $P'(x)=3 x^{2}+2 a x+b$, which has a root at approximately $r \approx -b / 2 a$, which is rather small compared to 100. Then $P(r) \approx -b^{2} / 4 a$. Assuming that this is greater than about $-100$, then the value of $c$ that produces the roots that are closest together is the closest integer to $-P(r)$ (the chance when this creates a double root is pretty rare). Let $-P(r)=c+s$, so that we can now assume that $s$ is uniformly distributed in $(-1 / 2,1 / 2)$. One can show that the difference between these roots is about $2 \sqrt{|s| / a}$. Since these roots are rather small, by Vieta's formulas the other root is near $-a$, so $M(P)$ is about $\frac{1}{2} \sqrt{a^{3} /|s|}$.

It's clear from this discussion that $a$ needs to be reasonably large for $M(P)$ to be large. Thus the condition $P(r)>-100$ is satisfied close to all the time - we will henceforth ignore it.

Set some $L$ and let's consider the expected number of $P$ so that $M(P)>L$. Then, for a given $a, b$, we need $|s|<a^{3} /(2 L)^{2}$. Summing over all $a$ and $b$, we find the probability is $2 \cdot 100 \cdot 100^{4} / 4 \cdot 1 /(2 L)^{2}$. Setting this equal to 20 gives us
$$L^{2}=\frac{10^{10}}{160} \Longrightarrow L \approx 7900$$
This is good enough for 14 points.