Calculating pH of weak base and strong acid: Why is K(a) not required?
Your doubt is legitimate because you are assuming that $K_{(a)}$ is a parameter that you can plug in every reaction. Actually $K_{(a)}$ is calculated using this equation in a solution of the acid with water:

$$
\ce{K_{(a)} = \frac{[H+][CN-]}{[HCN]}}
$$
That is derived from this ideal equilibrium: 
$$\ce{HCN ⇌ CN- + H+}$$

Even though it reflects the strength of that specific acid in a pure water solution it's practical usage is limited when the equilibrium is not perturbated by others factors.
Anything that will interfere with this equilibrium can cause problems to your assumptions.

In your specific case, the conditions are different there is an excess of $\ce{OH-}$ in the solution due to the strong base, so every $\ce{H+}$ produced will react with the $\ce{OH-}$. Due to the Le Chatelier principle the equilibrium will respond shifting to the right: dissociating more and more $\ce{HCN}$ from producing more $\ce{H+}$.

In simple words, if one reaction is favourable the solution doesn't care about your $K_{(a)}$ it will go following the most thermodynamic favourable path to the equilibrium.

Possible practical learning aim of this question?

So if you want to have high products amount (i.e. $\ce{CN-}$) and you have cheap reactant (i.e. $\ce{KOH}$) you can use an excess of reactant to shift the equilibrium where you want.

Bonus question: what can you do if you want to have HCN?