Ah! It's dangerous.

If the tire rotate $n$ times, then the board will move $n\times\text{circumference}$ relative to the tire. But the tire will move also. If the board move $l$ relative to the tire, then the tire will move $l$ also. So if the tire move $l$ , then the board will move $l+n\times\text{circumference}$ . but $l=\text{circumference}\times n$ . Therefore the board will move $2\times$ more than the tire. Therefore the solution is $2\times2m=4m$ .

For easy visualization, assume Peter starts at the end of the board exactly on top of the center of the wheel as shown in the top figure. As Peter pushes forward and the wheel rotates without slipping, the wheel moves relative to the ground $\text{ 2 m}$ forward. Because there is no slipping the wheel also moves relative to the board $\text{2 m}$ forward. Therefore, the distance the board moves forward relative to the ground is $2 + 2 = \text{4 m}$ . Since Peter is holding the board, Peter also moves forward $\boxed 4 \text{ m}$ .