Since the equation of the hyperbola is given by $$\frac {x^{2}}{9} - \frac {y^{2}}{16} = 1$$, we have semi-major axis $$a = 3$$ and semi-minor axis $$b = 4$$. The distance from the center to a focus can be calculated using $$c = \sqrt{a^2+b^2} = \sqrt{9+16} = 5$$.

According to the definition of a hyperbola, the absolute differences in distances from any point on a hyperbola to the foci is constant and equal to $$2a$$. Therefore, $$|PF_{1}| - |PF_{2}| = \pm2a = \pm6$$. Squaring both sides,
$$|PF_{1}|^2 - 2|PF_{1}|\cdot|PF_{2}| + |PF_{2}|^2 = 36$$.

This equation can be rearranged to give:
$$|PF_{1}|^2 + |PF_{2}|^2 = 36 + 2|PF_{1}|\cdot|PF_{2}|$$.

Since we are given that the angle ∠F1PF2 is 90°, it follows from the Pythagorean theorem that:
$$|PF_{1}|^2 + |PF_{2}|^2 = |F_{1}F_{2}|^2 = c^2 = 25$$.

Combining these two equations, we have:
$$100 = 36 + 2|PF_{1}|\cdot|PF_{2}|$$,
which simplifies to:
$$|PF_{1}|\cdot|PF_{2}| = 32$$.

The area of triangle ΔF1PF2 is given by one-half of the product of the lengths of the two segments forming the right angle, so:
$$S_{△F_{1}PF_{2}} = \frac{1}{2}(|PF_{1}|\cdot|PF_{2}|) = \frac{1}{2}(32) = 16$$.

Therefore, the area of the triangle is $\boxed{16}$.

The problem is essentially asking us to find the product of the distances from the point P to each focus and then calculate the area of the triangle using the fact that there's a right angle formed at P. The key in solving this problem is to use the definition of a hyperbola and apply the Pythagorean theorem effectively.