Let $$P$$ be a square matrix such that $$P^2 = I - P$$. For $$\alpha, \beta, \gamma, \delta \in \mathbb{N}$$, if $$P^\alpha + P^\beta = \gamma I - 29P$$ and $$P^\alpha - P^\beta = \delta I - 13P$$, then $$\alpha + \beta + \gamma - \delta$$ is equal to

We are given that $$P^2 = I - P$$, $$P^\alpha + P^\beta = \gamma I - 29P$$, and $$P^\alpha - P^\beta = \delta I - 13P$$, where $$\alpha, \beta, \gamma, \delta \in \mathbb{N}$$.

To express the powers of $$P$$ in the form $$aI + bP$$, observe that $$P^2 = I - P$$. Assuming $$P^{n-1} = a_{n-1}I + b_{n-1}P$$, one finds $$P^n = P\cdot P^{n-1} = a_{n-1}P + b_{n-1}P^2 = a_{n-1}P + b_{n-1}(I - P) = b_{n-1}I + (a_{n-1}-b_{n-1})P.$$

Computing these coefficients successively gives $$P^1 = 0\cdot I + 1\cdot P,\quad P^2 = 1\cdot I + (-1)\cdot P,\quad P^3 = (-1)I + 2P,\quad P^4 = 2I + (-3)P,\quad P^5 = (-3)I + 5P,\quad P^6 = 5I + (-8)P,\quad P^7 = (-8)I + 13P,\quad P^8 = 13I + (-21)P.$$

From $$P^\alpha + P^\beta = \gamma I - 29P$$ the coefficient of $$P$$ yields $$b_\alpha + b_\beta = -29$$, and from $$P^\alpha - P^\beta = \delta I - 13P$$ the coefficient of $$P$$ gives $$b_\alpha - b_\beta = -13$$. Solving these equations leads to $$b_\alpha = -21$$ and $$b_\beta = -8$$, which correspond to $$\alpha = 8$$ and $$\beta = 6$$.

The coefficients of $$I$$ in the same expressions are then $$\gamma = a_8 + a_6 = 13 + 5 = 18$$ and $$\delta = a_8 - a_6 = 13 - 5 = 8$$.

Hence, $$\alpha + \beta + \gamma - \delta = 8 + 6 + 18 - 8 = \boxed{24}$$. The correct answer is Option D.