Determination of solubility equilibrium using galvanic cell reactions
The leftmost and rightmost parts of the galvanic cell notation are supposed to be electrodes, not electrolytes (no KBr (aq) and the likes of it there). What is more, the double vertical line represents the interface where the junction potential has been eliminated ((c) Atkins), i.e. salt bridge, which in this case is the interface between electrolytes. The left hand side of the cell is anode, where the oxidation occurs, and the silver ions are reduced at the cathode, which is written on the right hand side. It seems to me that the correct notation is:

Ag (s) | AgBr (s) | KBr (aq) || AgNO3 (aq) | Ag (s)

In turn, the cell reaction is $\ce{Ag+ + Br- -&gt; AgBr}$, and the EMF of this reaction is +0.704 V. It is not surprising that the equilibrium constant of this reaction is much larger than unity, otherwise silver bromide wouldn't be so stable and so insoluble. What you're missing is that the solubility product, which you want to calculate, is an equilibrium constant of the inverse process - not the formation, but the dissolution of $\ce{AgBr}$. So take the inverse of the equilibrium constant which you obtained from the galvanic cell EMF, and you'll get $7.9\cdot10^{-13}$, which is pretty close to the reference values.