The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is:
[Given: The threshold frequency of platinum is $$1.3 \times 10^{15}$$ s$$^{-1}$$ and $$h = 6.6 \times 10^{-34}$$ Js]

Threshold frequency of platinum: $$\nu_0 = 1.3 \times 10^{15}$$ s$$^{-1}$$

Planck's constant: $$h = 6.6 \times 10^{-34}$$ Js

The minimum energy required equals the work function, which is the energy corresponding to the threshold frequency:

$$E = h\nu_0$$

$$E = 6.6 \times 10^{-34} \times 1.3 \times 10^{15}$$

$$E = 6.6 \times 1.3 \times 10^{-34+15}$$

$$E = 8.58 \times 10^{-19} \text{ J}$$

Hence, the correct answer is Option A.