I'll first take the dual of your statement by swapping all $\lor$'s with $\land$'s, and vice versa:
$$\left(a_1 \land \neg a_2 \land \neg a_3 \land ... \land \neg a_K \right) \lor$$
$$\left(\neg a_1 \land a_2 \land \neg a_3 \land ... \land \neg a_K \right) \lor$$
$$\left(\neg a_1 \land \neg a_2 \vee a_3 \land ... \land \neg a_K \right) \lor$$
$$...$$
$$\left(\neg a_1 \land \neg a_2 \land \neg a_3 \land ... \land a_K \right)
$$
Why did I do this? It is because this statement has an easy to understand interpretation: it says that exactly one of the variables is true.
Now, that statement can be paraphrased as saying that you want at least one of the variables to be true, but never two.
In logic, that would be:
$$(a_1 \lor a_2 \lor .... \lor a_K) \land$$
$$\neg (a_1 \land a_2) \land \neg (a_1 \land a_3) \land .... \land \neg (a_1 \land a_K) \land \neg (a_2 \land a_3) \land .... \land (a_{K-1} \land a _K)$$
Applying DeMorgan:
$$(a_1 \lor a_2 \lor ... \lor a_K) \land$$
$$(\neg a_1 \lor \neg a_2) \land (\neg a_1 \lor \neg a_3) \land ... \land ( \neg a_1 \lor \neg a_K) \land (\neg a_2 \lor \neg a_3) \land ... \land (\neg a_{K-1} \lor \neg a_K)$$
OK ... but how does that relate to your statement? Well, note that if my starting statement (in DNF) is the dual of your starting statement (in CNF), then the dual of my statement converted into CNF will be the dual of the conversion of your statement into DNF. So, we just need to get the dual of my end result, and that will be the DNF of your original statement. In other words, the DNF of your statement is:
$$(a_1 \land a_2 \land ... \land a_K) \lor$$
$$(\neg a_1 \land \neg a_2) \lor (\neg a_1 \land \neg a_3) \lor ...\lor (\neg a_1 \land \neg a_K) \lor (\neg a_2 \land \neg a_3) .... \lor (\neg a_{K-1} \land \neg a_K)$$
