BHSP is a rhombus whose diagonals intersect at point A. BA = 8 cm and AP=6 cm. The line AC is drawn parallel to BP as shown in the diagram below.

$\text{BP}^2 = \text{BA}^2 + \text{AP}^2$.

$$\text{BP}^2 = (8 \text{ cm})^2 + (6 \text{ cm})^2$$ $$\text{BP}^2 = 64 \text{ cm}^2 + 36 \text{ cm}^2$$ $$\text{BP}^2 = 100 \text{ cm}^2$$

$$\text{BP} = \sqrt{100 \text{ cm}^2}$$ $$\text{BP} = \boxed{10 \text{ cm}}$$

By the Midpoint Theorem, if a line passes through the midpoint of one side of a triangle and is parallel to another side, it bisects the third side and is half the length of the parallel side.

$$\text{AC} = \frac{1}{2} \times \text{BP}$$ $$\text{AC} = \frac{1}{2} \times 10 \text{ cm}$$ $$\text{AC} = \boxed{5 \text{ cm}}$$

Since diagonals bisect each other, $\text{BS} = 2 \times \text{BA}$ and $\text{HP} = 2 \times \text{AP}$.

$$\text{BS} = 2 \times 8 \text{ cm} = 16 \text{ cm}$$ $$\text{HP} = 2 \times 6 \text{ cm} = 12 \text{ cm}$$

$$\text{Area} = \frac{1}{2} \times \text{BS} \times \text{HP}$$ $$\text{Area} = \frac{1}{2} \times 16 \text{ cm} \times 12 \text{ cm}$$ $$\text{Area} = 8 \text{ cm} \times 12 \text{ cm}$$ $$\text{Area} = \boxed{96 \text{ cm}^2}$$