First of all, notice that you will only be on squares of one color. So we can simplify the board by taking only the squares of one color, and rotating the board diagonally.
Now, pieces move and jump horizontally and vertically.
Next, I will show a lower bound:
Say the king starts in a particular square. With only jumps, it can only visit every other column and every other row. That is, in the diagram below it can't leave the color it starts on.
Each of the colors is arranged the same way: eight dots, made up of a 2x3 rectangle and a single dot on each long side.
This arrangement only allows for nine total pieces to be jumped. So it cannot be done in one jump.
And now, the solution:
By the previous argument, it is not possible to do in one turn. It's also not possible to do in two: if one of the turns was a single step, you could just change the start or end position to make a one-turn solution. If both turns were entirely jumps, you could combine them into one turn.
But it is possible to do in three turns: a sequence of jumps, a step, and another sequence of jumps.
The black king starts on the left, then captures all the red pieces marked in black and ends on the space two cells right of its starting point. Then, it moves one step up, and its last turn captures all the red pieces marked in blue.