As Fedor Petrov says, this looks incorrect. The presence of

$$2/\pi = \prod_{n}\left(1-\frac{1}{(2n)^2}\right)$$

(known as the Wallis product) makes me think Sylvester is mistakenly comparing it to this somehow, with the missing implicit step being 'every prime is at least some even number', so $M\geq W$, since if $2n\leq p$ then we can replace the $(1-1/(2n)^2)$ factor with $(1-1/p^2)$ and only increase the product.

The error is, of course, that 'the greatest even number $\leq p$' is not unique for each $p$ - one gets the same factor appearing for both $2$ and $3$! If one corrects this by omitting $3$ then you get

$$ M > (1-1/9)\cdot\frac{2}{\pi} = \frac{16}{9\pi}.$$

I don't know if this correction is enough to salvage the remainder of Sylvester's argument. (Note that this is now consistent with the fact that $\prod_p (1-1/p^2)=6/\pi^2$).