The number of values of $$k$$, for which the system of equations :
$$(k+1)x + 8y = 4k$$
$$kx + (k+3)y = 3k - 1$$
has no solution, is :

We begin by writing the two given linear equations in the general form $$a_1x+b_1y=c_1,\;a_2x+b_2y=c_2.$$

For the first equation $$ (k+1)x+8y=4k$$ we can directly read

$$a_1=k+1,\qquad b_1=8,\qquad c_1=4k.$$

For the second equation $$kx+(k+3)y=3k-1$$ we obtain

$$a_2=k,\qquad b_2=k+3,\qquad c_2=3k-1.$$

A pair of simultaneous linear equations in two variables has no solution when the two lines are parallel but not coincident. In coefficient language this means that the ratios of the coefficients of $$x$$ and $$y$$ are equal, yet the same common ratio is not equal to the ratio of the constant terms. Symbolically, the condition for inconsistency is

$$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}.$$

We now impose this step by step. We first equate the two ratios of the coefficients of $$x$$ and $$y$$:

$$\frac{k+1}{k}=\frac{8}{k+3}.$$

To clear the denominators we cross-multiply:

$$ (k+1)(k+3)=8k. $$

Expanding the left-hand side gives

$$k^2+3k+k^2+3k= k^2+4k+3,$$

so the equation becomes

$$k^2+4k+3=8k.$$

Bringing all terms to the left yields

$$k^2+4k+3-8k=0\quad\Longrightarrow\quad k^2-4k+3=0.$$

This quadratic factors neatly:

$$k^2-4k+3=(k-1)(k-3)=0.$$

Hence the common-ratio condition produces the two candidate values

$$k=1\quad\text{or}\quad k=3.$$

We must now test each of these values in the remaining ratio so as to enforce the inequality part of the inconsistency criterion.

First we compute the ratio of the constant terms in general:

$$\frac{c_1}{c_2}=\frac{4k}{3k-1}.$$

Case $$k=1$$.

For $$k=1$$ we have

$$\frac{a_1}{a_2}=\frac{1+1}{1}=2,\qquad \frac{b_1}{b_2}=\frac{8}{1+3}=\frac{8}{4}=2,$$

so both coefficient ratios are equal to $$2$$ as required. The constant ratio becomes

$$\frac{c_1}{c_2}=\frac{4\cdot1}{3\cdot1-1}=\frac{4}{2}=2.$$

Here all three ratios are the same, which means the two equations represent the same line; the system therefore has infinitely many solutions, not zero solutions. Thus $$k=1$$ is rejected.

Case $$k=3$$.

For $$k=3$$ the coefficient ratios are

$$\frac{a_1}{a_2}=\frac{3+1}{3}=\frac{4}{3},\qquad \frac{b_1}{b_2}=\frac{8}{3+3}=\frac{8}{6}=\frac{4}{3},$$

so once again they are equal. The constant ratio now is

$$\frac{c_1}{c_2}=\frac{4\cdot3}{3\cdot3-1}=\frac{12}{8}=\frac{3}{2}.$$

We clearly see

$$\frac{4}{3}\neq\frac{3}{2},$$

which satisfies the essential inequality. Hence the two lines are parallel but distinct, producing no solution when $$k=3$$.

Since no other value of $$k$$ satisfies the condition for inconsistency, the system is without a solution for one and only one real value of $$k$$, namely $$k=3$$. Consequently, the number of such values is

$$1.$$

Hence, the correct answer is Option D.