Answer is A.
Effect of the above $3$ reversals for any $K$ is equivalent to left rotation of the array of size $n$ by $k$.
Let , $S[1\ldots7] = \begin{array}{|c|c|c|c|c|c|c|}\hline1&2&3&4&5&6&7\\ \hline\end{array}$
So, $n=7,k = 2$
reverse $(S,1,2)$ we get $[2,1,3,4,5,6,7]$
reverse $(S,3,7)$ we get $[2,1,7,6,5,4,3]$
reverse $(S,1,7)$ we get $[3,4,5,6,7,1,2]$
Hence, option $(A)$ rotates $s$ left by $k$ positions and is correct.