It is given that $f$ is function defined from domain $A$ to co-domain $B$, $f:A \rightarrow B$, and $f$ is surjective, in other and simple words, $\text{“every element of $B$ has pre-image or is mapped by some element(s) in $A$”}$. Mathematically this can be defined as $$\displaystyle \mathbf{\forall}_{y \ \in B} \ \mathcal{\exists}_{x \ \in A} \ y = f(x) \tag{1}$$ Since $f$ is function, so no element in $A$ can be left unmapped to some element in $B$.
Let,
$$\begin{align}A &= \{x_1, x_2, \dots, x_a, \dots, x_b, \dots, x_n\} \cr B &= \{y_1, y_2,\dots, y_n\}\end{align}$$
It is given that $\forall _{a_1, a_2 \in A} \ f(a_1) = f(a_2) \Rightarrow a_1R_Ea_2$, where $R_E$ is equivalence relation, so instead of $\sim$ I’m using that notation.
Now, according to defintion $(1)$, some elements of $A$ (say) $x_1,x_2,x_3$ would map to $y_i \in B$, similarly (say) $x_4, x_5$ would map to $y_j \in B$, hence sets of $x$ values forms a disjoint set while mapping to $y_i$ and $y_j$. So there are $|B|-1$ partitions or $|B|$ parts of $A$ and all of them are disjoint.
Let,
$$\begin{align}A_1 &= \{x_1, \dots, x_a\}; \ \forall _{x_i \in A_1} f(x_i) = y_1 \cr A_2 &= \{x_{a+1}, \dots, x_b\}; \ \forall _{x_i \in A_2} f(x_i) = y_2 \cr & \quad \vdots \cr A_n &= \{x_{m+1}, \dots, x_n\}; \ \forall _{x_i \in A_n} f(x_i) = y_n\end{align}$$
Let, $$\begin{align}& \ \ \ \ \ \forall_{a_1 \in A_1, a_2 \in A_2} \ f(a_1) \neq f(a_2) \Rightarrow a_1R_Ea_2 \cr &\equiv \forall_{a_1 \in A_1, a_2 \in A_2} \ False \Rightarrow a_1R_Ea_2 \cr &\equiv \forall_{a_1 \in A_1, a_2 \in A_2} \ True \cr &\equiv True\end{align}$$Although, whole first order logic expression is $True$ but it doesn't tell us about the truth value of $a_1R_Ea_2$, expression would be $True$ even if $a_1R_Ea_2$ or $a_1\cancel{R_E}a_2$, but we are actually concerned with $a_1R_Ea_2$. So lets’ make $a_1R_Ea_2 \equiv True$, then to make FOL expression $True$, $f(a_1)=f(a_2) \equiv True$. But this can only be $True$ when $a_1, a_2 \in A_i$. Using this we can say that $[a_j] = [a_k]: a_j,a_k \in A_i$, where $[.]$ is equivalence class.
So we get this,
$$\begin{align}\mathcal{Q} &= \{[x]: x \in A_j; 1 \leq j \leq n\}\cr &= \{[x]: x \in A_1 \cup A_2 \cup \dots \cup A_n\} \cr &= \{[x]: x \in A\}\end{align}$$
Now $\mathcal{Q}$ is similar to given set $\mathcal{E} = \{[x]: x \in A\}$ where $\mathcal{E}$ is set of all equivalance classes under $R_E$. New function $F$ has been defined as $F([x]) = f(x)$, and $F: \mathcal{Q} \rightarrow B$.
We can also write $\mathcal{Q}$ as, $$\mathcal{Q} = \{[x \in A_1], [x \in A_2], \dots, [x \in A_n]\} \tag{2}$$ $$\mathcal{Q} = \{\{x_1, x_2, \dots, x_a\}, \{x_{a+1}, \dots, x_b\}, \dots, \{x_{m+1}, \dots, x_n\}\} \tag{3}$$
If we look closely to $F$ then $F$ is taking $q \in \mathcal{Q}$ as an argument and we already know that $\exists_{y \in B}\forall_{x \in q} \ y = f(x)$ where $y$ is unique, hence $F(q)$ is unique. This shows that $\forall_{q_1, q_2 \in \mathcal{Q}} \ q_1 \neq q_2 \Rightarrow F(q_1) \neq F(q_2)$ which is definition of $\textbf{one-one}$ function.
In the beginning of the answer, we found the number of partitions sets or number of equivalence classes as $|B|$, this means $|\mathcal{Q}| = |B|$, here we can observe number of elements in domain is same as number of elements in co-domain and we already found $F$ to be one-one, so from here we can conclude $F$ is $\textbf{onto}$ function too.
$F$ is $\textbf{one-one}$ and $\textbf{onto}$, hence $F$ is $\textbf{bijective}$.
Since its MSQ type question,
$\textbf{(B), (C), (D)}$ are correct options.
