Why does the degree of dissociation change when we dilute a weak acid even though the equilibrium constant is constant?
See my comment above.

As a numerical example, take acetic acid ($\ce{AcOH}$), which has $K_\mathrm{a} = 1.8 \cdot 10^{-5}$.

This means that:

$$K_\mathrm{a} = \frac{[\ce{AcO-}][\ce{H+}]}{[\ce{AcOH}]}$$

And the total nominal concentration of acid is:

$$C_\mathrm{a} = [\ce{AcOH}] + [\ce{AcO-}]$$

Combining these two equations, you can see that the % of dissociated acid is:

$$\alpha = \frac{[\ce{AcO-}]}{C_\mathrm{a}} = \frac{K_\mathrm{a}}{K_\mathrm{a} + [\ce{H+}]}$$

i.e. not a constant, but a value that depends on the $\mathrm{pH}$. So no, there is no need for the 'degree of completion' of the reaction to be a constant.

You can even calculate the explicit concentrations of all the species.

Given that:

$$C_\mathrm{a} = \left(\frac{1}{[\ce{H+}]} + \frac{1}{K_\mathrm{a}}\right) \cdot ( [\ce{H+}]^2 - K_\mathrm{w}^2)$$

you can see that, for $\mathrm{pH} = 5$ you need:

$$C_\mathrm{a} = \left(\frac{1}{10^{-5}} + \frac{1}{1.8 \cdot 10^{-5}}\right) \cdot \left((10^{-5})^2 - (10^{-14})^2\right) = 1.5554 \cdot 10^{-5}$$

Then:

$$[\ce{AcOH}] = \frac{C_\mathrm{a} \cdot [\ce{H+}]}{K_\mathrm{a} + [\ce{H+}]} = 5.555 \cdot 10^{-6}$$

$$[\ce{AcO-}] = C_\mathrm{a} - [\ce{AcOH}] = 9.999 \cdot 10^{-6}$$

So the % of dissociated $\ce{AcOH}$ is:

$$\alpha = \frac{[\ce{AcO-}]}{C_\mathrm{a}} \approx 64 \%$$

Now repeat for $\mathrm{pH} = 3$:

$$C_\mathrm{a} = \left(\frac{1}{10^{-3}} + \frac{1}{1.8 \cdot 10^{-5}}\right) \cdot \left((10^{-3})^2 - (10^{-14})^2\right) \approx 0.05656$$

$$[\ce{AcOH}] = \frac{C_\mathrm{a} \cdot [\ce{H+}]}{K_\mathrm{a} + [\ce{H+}]} \approx 0.05556$$

$$[\ce{AcO-}] = C_\mathrm{a} - [\ce{AcOH}] \approx 0.001$$

$$\alpha = \frac{[\ce{AcO-}]}{C_\mathrm{a}} \approx 1.8 \%$$

So you see, in a more dilute mixture the $\mathrm{pH}$ is higher, and the dissociation reaction of acetic acid is 'more complete', whereas at higher concentration the pH is lower, and there is proportionally less acetate anion.

All of these systems are at equilibrium.