Let $L$ be a regular language over $\Sigma=\{a,b,c\}$. Build a finite automaton for $L/\{a\}$.
Because $L$ is regular then a DFA exists for it: $A=(\Sigma, Q, q_0, F, \delta)$.
Let $M$ be a finite automaton, $L(M)=L/\{a\}$.
$M=(\Sigma, Q\times\{0,1\}, (q_0,0), \delta', F\times\{1\})$ with the transition function defined below for all $q\in Q, \sigma \in \Sigma$: $$ \delta'((q,0),\sigma)=(\delta(q,\sigma),0)\\ \delta'((q,0),\epsilon)=(\delta(q,a),1) $$
The reasoning is that in state $0$ we just read letters in $M$ because those exact letters also exist in $L$. But when we reach end of input in $M$ via $\epsilon$ we still need to read $a$ in $L$. I'm not sure if it's valid to assume that $\epsilon$ means end of input?