Katz/Lindell Problem 2.2
You neglected the most important part, I believe; the encryption map itself may be probabilistic which means we might have $$c \leftarrow \mathrm{Enc}_k(m),$$ instead of $$c =\mathrm{Enc}_k(m),$$ which is deterministic. In the first case a second encryption with the same key and message might result in a different ciphertext. Once you realise this and given that the key space is finite consider for each fixed key message pair $(k,m)$ the distribution of the random variable $$\mathrm{Enc}_k(m) $$ over $\mathcal{C}$ and note that this distribution where each probability is a real number can be approximated arbitrarily closely by a rational distribution with rational probabilities.

Let us say $p_i=a_i/b_i,$ is the rational arbitrarily accurate approximate probability for the ciphertext $c_i$ belonging to the set of texts that appear with nonzero probability for the given $(k,m).$ Rewriting the probabilities as fractions with denominator $\mathrm{LCM}(b_i)$ we can now replace the key $k$ with $$\mathrm{LCM}(b_i) a_i/b_i$$ keys all of which will now map deterministically to the single ciphertext $c_i.$ 

Proceeding this way for each $(k,m)$ pair we obtain a deterministic scheme.