a. The sides of the initial triangle are distributed between the pieces. As $k \ge 2$, the sides of the pieces must also pass through the interior of the initial triangle. So the sum of the perimeters of the $k$ pieces is greater than $P$. Hence there must exist a piece with perimeter greater than $\frac{P}{k}$.

b. Let the area of the equilateral triangle be $S$. Then there must exist a piece $\Delta$ with area at least $\frac{S}{k}$. If $\Delta$ were equilateral, it would be similar to the initial triangle with a scale factor of $\frac{1}{\sqrt{k}}$, so its perimeter would be $\frac{P}{\sqrt{k}}$. However, among triangles with a fixed area, an equilateral triangle has the smallest perimeter. Therefore the perimeter of $\Delta$ is at least $\frac{P}{\sqrt{k}}$.