Solution:

The answer is (C). Let $ABCD$ be a quadrilateral, and suppose that the length of $AB$ is $1$, that of $BC$ is $7$, and that $CD = AD = 5$. Consider the triangle $ABC$ and the triangle $CDA$; let $AH$ be the altitude of $ABC$ relative to $BC$, and $AK$ the altitude of $CDA$ relative to $CD$. We have $AH \leq AB$ (note the right triangle $ABH$ of which $AB$ is the hypotenuse), and similarly $AK \geq AD$.

By Ptolemy's theorem we have
$$\frac{1}{2}(BC \cdot AH + CD \cdot AK) \leq \frac{1}{2}(BC \cdot AB + CD \cdot AD) = \frac{1}{2}(7 + 25) = 16.$$
On the other hand, there exists a quadrilateral satisfying the requirements, obtained by joining the hypotenuses of two right triangles, one with legs of length $1$ and $7$, the other with both legs of length $5$: this is because the hypotenuse of the first has length $\sqrt{1 + 7^{2}} = \sqrt{50}$, and that of the second has length $\sqrt{5^{2} + 5^{2}}$, that is, again $\sqrt{50}$. There is therefore a quadrilateral $ABCD$ as described above, in which the angles at $B$ and at $D$ are right angles, whose area is exactly $16$.

Alternatively, one can observe that if we consider the triangles formed by adjacent sides and the diagonal, they have maximum area when they are right triangles. But if we take the right triangle with legs $1$ and $7$ and the one with legs $5$ and $5$, they have hypotenuses of the same length, so we can glue them together along the diagonal to obtain the quadrilateral of maximum area, equal to $7/2 + 25/2 = 16$.