What does enzymatic equilibrium in % represent?
You are correct in stating the relative rates of the two reactions.  What may have you confused is the percent sign in the statement.  The percent sign is not a unit in the sense that you are taking it.

What you have in a simple reaction like you are examining are two opposing rates of reaction.  The rate of a catalytic reaction, like others, usually depends on the concentrations of at least some of the reactants in the direction under consideration.

At equilibrium, by definition, the concentration of every material involved in the reaction is unvarying.  Otherwise, the reaction would not have reached equilibrium yet.  What that also means is that the rates of the forward &amp; inverse reactions are equal at equilibrium.  So what does the statement signify?   It means that thermodynamically, the synthase reaction is the more unfavorable direction.  That is, at equilibrium, the product concentration of the synthase direction will be in the minority. 

The statement involving the percent sign is just giving the ratio between the [arbitrary] forward direction, here stated as the synthase direction, and the inverse reaction.  The percent sign is just a mathematically short way of stating that the number stated is parts per hundred.

The equilibrium constant K is exactly the same ratio, stated as a decimal value rather than a integer per hundred.