Solution:

The answer is $\mathbf{(C)}$. Let us first observe that the height of the trapezoid $ABCD$ equals $3$: indeed, letting $H$ be the foot of the altitude drawn from $C$ to $AB$, the triangle $HBC$ is right-angled at $H$ and we have the equalities $BC=5$ and $HB=\frac{1}{2}(AB-CD)=4$, from which (by the Pythagorean theorem) it follows that $HC=3$ as stated.

The given polygon can be decomposed into the union of four squares (having total area $5^2+5^2+15^2+7^2=324$), the original trapezoid (having area $\frac{(15+7) \cdot 3}{2}=33$) and four triangles, whose total area is exactly equal to twice the area of the trapezoid (that is, $66$).

The reason for this last statement is that the triangle determined by two segments of lengths $\ell_1, \ell_2$ forming an angle $\theta$ between them has the same area as a triangle determined by two segments of lengths $\ell_1, \ell_2$ forming an angle $180^\circ-\theta$ between them: applying this fact repeatedly, we find that the following pairs of triangles have the same area: $AA_1A_2$ and $ABD$, $BB_1B_2$ and $ABC$, $CC_1C_2$ and $BCD$, $DD_1D_2$ and $ADC$. Since the sum of the areas of triangles $ABD$ and $BCD$ is the area of the trapezoid, and the same holds for the areas of $ADC$ and $ABC$, we obtain, as desired, that the areas of triangles $AA_1A_2$, $BB_1B_2$, $CC_1C_2$, $DD_1D_2$ sum to twice the area of the trapezoid.

The correct answer is therefore $324+33+66=423$.