Drawn DFA is not correct, its still NFA

$q_0$ will not contain the transition $\textbf{a}$ to itself and $q_1$ will contain the self loop $\textbf{a}$ to itself

$q_0$ will not contain the transition $\textbf{a}$ to itself and $q_1$ will contain the self loop $\textbf{a}$ to itself