There's a systematic approach to this problem. Let $L_1$ and $L_2$ are the two languages of odd number of $b$'s and even number of $a$'s respectively.
$\therefore \mathrm{DFA}(L_1)$ is below.
$\therefore \mathrm{DFA}(L_2)$ is below.
We need to determine $L=(L_1 \cup L_2)^{*}$
So, the states of $\mathrm{DFA}(L)$ will be the cross product of the states of $\mathrm{DFA}(L_1)$ and $\mathrm{DFA}(L_2)$. In this automation, the pair containing the initial states of both $\mathrm{DFA}(L_1)$ and $\mathrm{DFA}(L_2)$ will be the initial state and the pair containing final (accepting) states of both of them will be final (accepting) state. Besides all other pair of states will preserve the edges of $\mathrm{DFA}(L_1)$ and $\mathrm{DFA}(L_2)$. So there will be 4 pair of states: $(13)$, $(14)$, $(23)$ and $(23)$.
$(13)$ will be the initial state as both $1$ and $3$ are the initial states of $\mathrm{DFA}(L_1)$ and $\mathrm{DFA}(L_2)$ respectively. Thus $(23)$ will be final (accepting) state.
$\therefore \mathrm{DFA}(L)$ is below.