Four bridge hands with no two people having 8 or more cards of the same suit between them.

First, your two choices for the denominator are the same since $${52\choose13}{39\choose13}{26\choose13}=\frac{52!}{39!13!}\frac{39!}{26!13!}\frac{26!}{13!13!}=\frac{52!}{(13!)^4}$$ Second, you are right that everyone needs to have a 4333 hand, but there are many more than $4!$ ways to do it. We will make 4 hands with a specific 4 cards suit, then distribute the hands to the four players.

North hand with 4 spades $${13\choose4}{13\choose3}{13\choose3}{13\choose3}$$ East hand with 4 hearts, from remaining card. $${9\choose3}{10\choose4}{10\choose3}{10\choose3}$$ South hand with 4 diamonds, from remaining cards $${6\choose3}{6\choose3}{7\choose4}{7\choose3}$$ Finally, West hand with 4 clubs, from remaining cards $${3\choose3}{3\choose3}{3\choose3}{4\choose4}$$ The number of ways that no team has a 8 cards suit between them is $$4!{13\choose4}{13\choose3}{13\choose3}{13\choose3}{9\choose3}{10\choose4}{10\choose3}{10\choose3}{6\choose3}{6\choose3}{7\choose4}{7\choose3}{3\choose3}{3\choose3}{3\choose3}{4\choose4}$$ which could be simplify to $$4!\left(\frac{13!}{4!(3!)^3}\right)^4$$ We could have found this expression directly. The interior of the parenthesis is the ways to distribute a suit evenly (to the fourth power for the four suits) and the $4!$ in front to decide who received each 4 cards suits.

The probability is $$4!\left(\frac{13!}{4!(3!)^3}\right)^4\times\frac{(13!)^4}{52!}\approx0.000\ 931\ 419\ldots$$