let $C_{1} = \{a_{11}...a_{i1}\}$ and $C_{2} = \{a_{12}...a_{j2}\}$ also, let $C'_{1} = \{a'_{11}...a'_{i1}\}$ and $C'_{2} = \{a'_{12}...a'_{j2}\}$. Where $|C_1| = |C'_1| = i$ and $|C_2| = |C'_2| = j$
Now, $\forall (a_{p1},a_{q2}): (a_{p1},a_{q2})\in C_1 \times C_2$, let there be a function $f:C_1 \times C_2 \rightarrow \{l_1,l_2...l_m\} $, where $\{l_1,l_2...l_m\}$ are some labels on each tuple.
Similarly, $\forall (a'_{p1},a'_{q2}): (a'_{p1},a'_{q2})\in C'_1 \times C'_2$,let there be a function $f':C'_1 \times C'_2 \rightarrow \{l_1,l_2...l_m\} $, where $\{l_1,l_2...l_m\}$ are same labels as in the previous function.
Furthermore, there is exactly the same numbers of tuples labeled $l_i$ in $C_1 \times C_2$ as there are in $C'_1 \times C'_2$, forall $i \in \{1,..,m\}$.
Now, can we prove that there exists a bijection, B:$C_1 \cup C_2 \rightarrow C'_1 \cup C'_2$ such that $f((B(a_{p1}),B(a_{q2}))) = f((a_{p1},a_{q2})) $ ?
Also, $C_1 \cap C_2= \phi$ and $C'_1 \cap C'_2= \phi$
Suppose two sets $C_1$ and $C_1'$ have the same cardinality. Similarly the two sets $C_2$ and $C_2'$ have the same cardinality. If that cardinality is finite, then we can enumerate each pair of sets with an index set: $I_1 = \left\{1,\dots,m\right\}$ and $I_2 = \left\{1, \dots, n\right\}$. Explicitly these enumerations are provided as:
$$ \begin{aligned} C_1 &= \left\{ {}_1c_1, \dots, {}_1c_m \right\} \\ C_1' &= \left\{ {}_1c_1', \dots, {}_1c_m' \right\} \end{aligned} $$
and:
$$ \begin{aligned} C_2 &= \left\{ {}_2c_1, \dots, {}_2c_n \right\} \\ C_2' &= \left\{ {}_2c_1', \dots, {}_2c_n' \right\} \end{aligned} $$
Suppose these sets are mutually disjoint. Then the following is a bijection, $B: C_1\cup C_2 \rightarrow C_1'\cup C_2'$:
$$ B(\alpha) = \left\{ \begin{aligned} B_1(\alpha), & \,\,\alpha \in C_1 \\ B_2(\alpha), & \,\,\alpha \in C_2 \\ \end{aligned}\right. $$
where we define $B_1: C_1 \rightarrow C_1'$ and $B_2: C_2 \rightarrow C_2'$ as:
$$ B_1({}_1c_j) = {}_1c_j', \quad \forall j \in I_1 $$
and
$$ B_1({}_2c_k) = {}_2c_k', \quad \forall k \in I_2 $$
Let's prove that this is a bijection. First we'll show it's onto. Choose an element $\beta \in C_1' \cup C_2'$. Necessarily, $\beta \in C_1' \oplus \beta \in C_2'$. If $\beta \in C_1'$ then $\beta = {}_1c_j'$ for some $j\in I_2$. We know that ${}_1c_j \in C_1$, and it follows that $B_1({}_1c_j) = {}_1c_j'$. If $\beta \in C_2'$, the same argument can be applied. We conclude $B$ is onto.
Now let's show that it is one-to-one. Suppose there are two elements $\alpha, \gamma \in C_1 \cup C_2$ such that $B(\alpha) = B(\gamma) = \beta$, but $\alpha \neq \gamma$. Then there is a unique ${}_1c_k' \in C_1'$ or ${}_2c_k' \in C_2$ such that $\beta = {}_1c_k'$ or $\beta = {}_2c_k'$. Without loss of generalization we consider the first case, $\beta = {}_1c_k'$ It follows that $\exists {}_1c_j, {}_1c_\ell \in C_1$ such that ${}_1c_j = \alpha$ and ${}_1c_\ell = \beta$. But:
$$ B({}_1c_j) = B({}_1c_\ell) = {}_1c_k' \Longrightarrow j = \ell = k \Longrightarrow \alpha = {}_1c_k = \beta $$
which is a contradiction. We apply the came argument to the case $\beta \in C_2'$ and conclude that $B$ is one-to-one.
We have shown that $B$ is one-to-one and onto, so by definition, $B$ is a bijection.
If you show that $f$ and $f'$ are bijections, then you can show that the statement I think you want to prove can be written as the composition of bijections, which is necessarily a bijection. However, as I pointed out in the comments, it seems there are some mistakes in the question statement.