Lets take example and evaluate the query.
$$\textbf{r1}\\\begin{array}{|l|l|l|l|} \hline A & B & C \\ \hline 1 & a & \text{Raju} \\ \hline 2 & a & \text{Rani} \\ \hline 3 & b & \text{Rakesh} \\ \hline 4 & b & \text{Vani} \\ \hline \end{array} \\ \textbf{r2}\\\begin{array}{|l|l|l|l|} \hline A & B & C \\ \hline 1 & a & \text{Raju} \\ \hline 2 & a & \text{Rani} \\ \hline 3 & b & \text{Rakesh} \\ \hline 4 & b & \text{Vani} \\ \hline \end{array}$$
When we perform $\Pi_{AB}(r1),$ we get $$\textbf{r1}\\\begin{array}{|l|l|l|l|} \hline A & B \\ \hline 1 & a \\ \hline 2 & a \\ \hline 3 & b \\ \hline 4 & b \\ \hline \end{array}$$
And when we perform $\Pi_{BC}(r2),$ we get $$\textbf{r2}\\\begin{array}{|l|l|l|l|} \hline B & C \\ \hline a & \text{Raju} \\ \hline a & \text{Rani} \\ \hline b & \text{Rakesh} \\ \hline b & \text{Vani} \\ \hline \end{array}$$
Now, perform natural join, so $r1.B=r2.B$ will be combine.
$\Pi_{AB}(r1)\Join \Pi_{BC}(r2)$, we get $$\textbf{r1}\\\begin{array}{|l|l|l|l|} \hline A & B & C \\ \hline 1 & a & \text{Raju} \\ \hline 1 & a & \text{Rani} \\ \hline 2 & a & \text{Raju} \\ \hline 2 & a & \text{Rani} \\ \hline 3 & b & \text{Rakesh} \\ \hline 3 & b & \text{Vani} \\ \hline 4 & b & \text{Rakesh} \\ \hline 4 & b & \text{Vani} \\ \hline \end{array} $$
Same thing we can do in SQL
select r1.A, r2.B, r2.C from r1, r2 where r1.B = r2.B
Here we can't select C from r1$($we project C only on r2$).$
So, the correct answer is $(A)$.