If $$a_1, a_2, a_3, \ldots$$ are in A.P. such that $$a_1 + a_7 + a_{16} = 40$$, then the sum of the first 15 terms of this A.P is:

Let us denote the first term of the A.P. by $$a_1$$ and its common difference by $$d$$. The general $$n^{\text{th}}$$ term of an arithmetic progression is given by the formula

$$a_n \;=\; a_1 + (n-1)d.$$

We are told that $$a_1 + a_7 + a_{16} = 40$$. Using the general-term formula for each of these three terms, we write

$$\begin{aligned} a_1 &= a_1,\\ a_7 &= a_1 + (7-1)d = a_1 + 6d,\\ a_{16} &= a_1 + (16-1)d = a_1 + 15d. \end{aligned}$$

Adding them exactly as given, we have

$$a_1 + a_7 + a_{16} \;=\; a_1 + (a_1 + 6d) + (a_1 + 15d).$$

Collecting like terms,

$$3a_1 + 21d = 40.$$

Factoring out the common factor $$3$$ from the left hand side,

$$3\bigl(a_1 + 7d\bigr) = 40.$$

Dividing both sides by $$3$$ gives

$$a_1 + 7d = \frac{40}{3}.$$

We now turn to the required quantity: the sum of the first 15 terms. The sum of the first $$n$$ terms of an A.P. is given by

$$S_n \;=\; \frac{n}{2}\,\bigl[\,2a_1 + (n-1)d\,\bigr].$$

Substituting $$n = 15$$, we get

$$S_{15} \;=\; \frac{15}{2}\,\bigl[\,2a_1 + 14d\,\bigr].$$

Notice that $$2a_1 + 14d = 2\bigl(a_1 + 7d\bigr).$$ We have already found $$a_1 + 7d = \dfrac{40}{3}$$, so

$$2a_1 + 14d = 2 \times \frac{40}{3} = \frac{80}{3}.$$

Substituting this back into the expression for $$S_{15}$$:

$$\begin{aligned} S_{15} &= \frac{15}{2} \times \frac{80}{3}\\[6pt] &= \frac{15 \times 80}{6}\\[6pt] &= \frac{1200}{6}\\[6pt] &= 200. \end{aligned}$$

So the sum of the first 15 terms of the given A.P. is $$200$$.

Hence, the correct answer is Option D.