Assume that the input is a valid description of a DFA $W$ (for any reasonable description this is a decidable problem).
You need to show that, given $W$, is is possible to decide whether there exists a word $w \in \{b,c\}^*$ that is accepted by $W$.
Since $W$ is a DFA, $L(W)$ is a regular language. By the closure properties of regular languages, $L' = L(W) \cap \{ b,c \}^*$ is also regular (and you can construct a DFA $W'$ for it by simply deleting all transitions labelled $a$ from $W$).
Then the problem is equivalent to deciding whether $L'$ is empty. This problem is known to be decidable. In fact, you can even solve it in linear time w.r.t. the natural description of the DFA: you just need to decide whether there exists a path from the initial state to a final state in the graph induced by $W'$ (which can be done using, e.g., a breadth first search).
Here is an alternative (slower) approach that more closely resembles the one in your question. Design a Turing machine $T$ that, on input $W$, behaves as follows:
- Let $n$ be the number of states of $W$.
- For each $i=0, \dots, n-1$:
- For each word $w \in \{b,c\}^i$:
- Simulate $W$ on $w$.
- If $W$ accepted $w$ then accept.
- Reject
It is clear that $T$ always halts and that if $W \not\in K$ then $T$ must reject. Therefore we only need to show that if $W \in K$ then $T$ accepts.
To show the above, it suffices to prove that the existence of some word $w \in \{b,c\}^*$ accepted by $W$ implies the existence of some $w' \in \{b,c\}^*$ with $|w'| \le n−1$ that is also accepted by $W$.
Among all possible words $w \in \{b,c\}^*$ accepted by $W$, let $w'$ be one of those minimizing $|w|$ and suppose towards a contradiction that $|w'| \ge n$. Since the pumping length of $L$ is at most $n$, we can invoke the pumping lemma to write $w'$ as $xyz$ with $|y| \ge 1$. Then, $w''=xz \in L$ belongs to $\{b,c\}^∗$, is accepted by $W$, and satisfies $|w''| < |w'|$. This contradicts the choice of $w'$.
