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 trimino 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 $B$ 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 marks the squares 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 $k=4$.