Solution:

We will show that player $A$ wins if $k=1,2$ or $3$, but player $B$ wins if $k=4$. Thus the smallest $k$ for which $B$ has a winning strategy exists and is equal to $4$.

If $k=1$, player $A$ marks the upper left corner of the square and then fills it as follows.

If $k=2$, player $A$ marks the upper left corner of the square. Whatever square player $B$ marks, then player $A$ can fill in the square in exactly the same pattern as above except that he doesn't put the triomino which covers the marked square of $B$. Player $A$ wins because he has left only two unmarked squares uncovered.

For $k=3$, player $A$ wins by following the same strategy. When he has to mark a square for the second time, he marks any yet unmarked square of the triomino that covers the marked square of $B$.

Let us now show that for $k=4$ player $B$ has a winning strategy. Since there will be $21$ unmarked squares, player $A$ will need to cover all of them with seven $L$-shaped triominoes. We can assume that in his first move, player $A$ does not mark any square in the bottom two rows of the chessboard (otherwise just rotate the chessboard). In his first move player $B$ marks the square labeled $1$ in the following figure.

If player $A$ in his next move does not mark any of the squares labeled $2,3$ and $4$ then player $B$ marks the square labeled $3$. Player $B$ wins as the square labeled $2$ is left unmarked but cannot be covered with an $L$-shaped triomino.

If player $A$ in his next move marks the square labeled $2$, then player $B$ marks the square labeled $5$. Player $B$ wins as the square labeled $3$ is left unmarked but cannot be covered with an $L$-shaped triomino.

Finally, if player $A$ in his next move marks one of the squares labeled $3$ or $4$, player $B$ marks the other of these two squares. Player $B$ wins as the square labeled $2$ is left unmarked but cannot be covered with an $L$-shaped triomino.

Since we have covered all possible cases, player $B$ wins when $k=4$.