How to estimate the equilibrium constant of the reaction between NaF and HCl
We can entirely ignore the sodium cation; it is only a spectator ion. This means, the reaction we are observing is:

$$\ce{HCl + F- &lt;=&gt; Cl- + HF}\tag{1}$$

The equilibrium constant for this reaction is:

$$K = \frac{[\ce{Cl-}][\ce{HF}]}{[\ce{HCl}][\ce{F-}]}\tag{2}$$

We can expand this equation:

$$K = \frac{[\ce{Cl-}][\ce{H+}][\ce{HF}]}{[\ce{HCl}][\ce{H+}][\ce{F-}]}\tag{2'}$$

And then we realise that that is nothing else than:

$$K = \frac{[\ce{Cl-}][\ce{H+}][\ce{HF}]}{[\ce{HCl}][\ce{H+}][\ce{F-}]} = \frac{[\ce{Cl-}][\ce{H+}]}{[\ce{HCl}]} \cdot \frac{[\ce{HF}]}{[\ce{H+}][\ce{F-}]} = \frac{K_\mathrm{a}(\ce{HCl})}{K_\mathrm{a}(\ce{HF})} \tag{2''}$$

So the equilibrium constant is the fraction of the acidity constants of $\ce{HCl}$ and $\ce{HF}$. We know that $\ce{HCl}$ is a much stronger acid than $\ce{HF}$. This is reflected by the inequation:

$$K_\mathrm{a}(\ce{HCl}) &gt; K_\mathrm{a}(\ce{HF}) \tag{3}$$

Since the numerator is larger than the denominator, the value of the fraction must be larger than $1$. Therefore, the product side is preferred.

If your book arrives at any other conclusion, it is disregarding experimental results and should be replaced.