Unlikely

$M$ and $N$ are coprime iff they have no common prime factors.

Thus $M$ and $N$ are not both divisible by 2; not both divisible by 3; not both divisible by 5; and so on. This gives $p = \left(1-\frac 1{2^2}\right)\cdot\left(1-\frac 1{3^2}\right)\cdot\left(1-\frac 1{5^2}\right)\cdots\left(1-\frac 1{p^2}\right)\cdots$

If we expand this fraction, we get $p = 1 - \frac1{2^2}-\frac1{3^2}-\frac1{5^2}+\frac1{6^2}-\frac1{7^2}+\frac1{10^2}-\frac1{11^2}-\frac1{13^2}+\frac 1{14^2}+ \frac 1{15^2}+-\cdots \\ = \sum_{n=1}^\infty \frac{\mu(n)}{n^2},$ where $\mu(n)$ is the Moebius function, which is 0 for any number divisible by a square, and $(-1)^r$ for any number containing $r$ distinct prime factors.

To evaluate $p$ , let's try to find its multiplicative inverse. We will add multiples of $p$ to cancel the terms one by one, leaving only 1, as follows: $\begin{array}{rccccccccccc} p = & 1 & -1/2^2 & - 1/3^2 & & -1/5^2 & +1/6^2 & -1/7^2 & & & +1/10^2 & +\cdots \\ p/2^2 = & & +1/2^2 & & -1/4^2 & & -1/6^2 & & & & -1/10^2 & + \cdots \\ p/3^2 = & & & +1/3^2 & & & -1/6^2 & & & -1/9^2 & & + \cdots \\ p/4^2 = & & & & +1/4^2 & & & & -1/8^2 & & & +\cdots \\ p/5^2 = & & & & & +1/5^2 & & & & & -1/10^2 & +\cdots \\ p/6^2 = & & & & & & +1/6^2 & & & & & + \cdots \\ p/7^2 = & & & & & & & +1/7^2 & & & & + \cdots \\ p/8^2 = & & & & & & & & +1/8^2 & & & +\cdots \\ p/9^2 = & & & & & & & & & +1/9^2 & & + \cdots \\ p/10^2 = & & & & & & & & & & +1/10^2 & + \cdots \\ \vdots = & \vdots & & & & & & & & & \vdots & + \ddots \\ \hline \text{sum} = & 1 & & & & & & & & & & \end{array}$

I leave it to the reader to convince him/herself that in this way all terms cancel, except the first. Thus we have $p\left(1 + \frac1{2^2} + \frac1{3^2} + \frac1{4^2} + \cdots\right) = 1.$ The series in brackets is well-known to converge to $\pi^2/6$ , so that $p = \frac 1 {\sum_{n=1}^\infty n^{-2}} = \frac 1{\pi^2/6} = \frac 6{\pi^2}\approx 0.\boxed{6079}27102.$

Note : In general, $\sum_{n=1}^\infty \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)},$ with $\zeta(n) = \sum_{p\ \text{prime}} (1-p^{-s})$ the Riemann zeta function. Here $s = 2$ and $\zeta(s) = \pi^2/6$ .