Answer
The minimum pressing force each hand must exert is $38.75N$
Work Step by Step
To keep the book from falling, maximum static friction $f_s^{max}$ has to be at least as big as the book's weight. In other words,
$$f_s^{max}\ge mg$$ $$\mu_s F_N^{min}=mg$$
$F_N$ is perpendicular with $f_s$, so in here, $F_N$ is the total pressing force 2 hands exert on the cover of the book.
If we call the pressing force each hand exerts on the cover $p$, since there are 2 hands and we assume they press with equal force, we have $F_N=2p$
Therefore, $$2\mu_s p_{min}=mg$$ $$p_{min}=\frac{mg}{2\mu_s}=\frac{31}{2\times0.4}=38.75N$$
