Let $$x = \left(8\sqrt{3} + 13\right)^{13}$$ and $$y = \left(7\sqrt{2} + 9\right)^{9}$$. If $$[t]$$ denotes the greatest integer $$\leq t$$, then

Write the conjugate numbers
 $$x' = \left(13 - 8\sqrt{3}\right)^{13}, \qquad y' = \left(9 - 7\sqrt{2}\right)^{9}$$

Case 1: Sign and size of the conjugates.

Evaluate the bases approximately:
 $$13 - 8\sqrt{3} \; \approx \; 13 - 8(1.732) \;=\; 13 - 13.856 \;=\; -0.856,$$
 $$9 - 7\sqrt{2} \;\approx\; 9 - 7(1.414) \;=\; 9 - 9.898 \;=\; -0.898.$$
Hence $$-1 \lt 13 - 8\sqrt{3} \lt 0$$ and $$-1 \lt 9 - 7\sqrt{2} \lt 0.$$
Raising numbers strictly between $$-1$$ and $$0$$ to positive integer powers keeps the results between $$-1$$ and $$0$$. Therefore
 $$-1 \lt x' \lt 0, \qquad -1 \lt y' \lt 0.$$

Case 2: Form an integer by adding each number to its conjugate.

Expand using the binomial theorem. Because odd-power terms cancel, only even-power terms survive:

$$\left(13 + 8\sqrt{3}\right)^{13} + \left(13 - 8\sqrt{3}\right)^{13}$$
 $$= 2 \sum_{m=0}^{6} \binom{13}{2m}\, 13^{\,13-2m}\,\left(8\sqrt{3}\right)^{2m},$$
 $$= 2 \sum_{m=0}^{6} \binom{13}{2m}\, 13^{\,13-2m}\,8^{\,2m}\,3^{\,m}.$$
All factors are integers, and the additional factor $$2$$ makes the whole sum even. Call this even integer $$N_1$$.

Similarly,

$$\left(9 + 7\sqrt{2}\right)^{9} + \left(9 - 7\sqrt{2}\right)^{9}$$
 $$= 2 \sum_{m=0}^{4} \binom{9}{2m}\, 9^{\,9-2m}\,\left(7\sqrt{2}\right)^{2m},$$
which is also an even integer. Call it $$N_2$$.

Case 3: Locate $$x$$ and $$y$$ with respect to these even integers.

Because $$x'$$ and $$y'$$ are negative,

$$x = \left(13 + 8\sqrt{3}\right)^{13} = N_1 - x' = N_1 + |x'|,$$
 and $$0 \lt |x'| \lt 1.$$ Hence $$N_1 \lt x \lt N_1 + 1,$$ so the greatest integer not exceeding $$x$$ is
 $$[x] = N_1.$$

Exactly the same argument gives
 $$[y] = N_2,$$
with $$N_2$$ even.

Case 4: Parity of the required quantities.

Both $$N_1$$ and $$N_2$$ are even, so
 $$[x] \text{ is even}, \quad [y] \text{ is even}, \quad [x] + [y] \text{ is even}.$$

The only option matching these facts is Option A.

Final answer: Option A.