If the perpendicular bisector of the line segment joining the points $$P(1, 4)$$ and $$Q(k, 3)$$ has $$y$$-intercept equal to $$-4$$, then a value of $$k$$ is:

We have two points $$P(1,\,4)$$ and $$Q(k,\,3)$$. Our task is to use the given condition on the perpendicular bisector of $$\overline{PQ}$$ to find the unknown $$k$$.

First, recall the midpoint formula. If $$P(x_1,\,y_1)$$ and $$Q(x_2,\,y_2)$$ are the endpoints of a segment, then the midpoint $$M$$ has coordinates

$$M\Bigl(\,\dfrac{x_1+x_2}{2},\;\dfrac{y_1+y_2}{2}\Bigr).$$

Substituting $$x_1=1,\;y_1=4,\;x_2=k,\;y_2=3$$ we obtain

$$M\Bigl(\dfrac{1+k}{2},\;\dfrac{4+3}{2}\Bigr)=\Bigl(\dfrac{1+k}{2},\;\dfrac{7}{2}\Bigr).$$

Next, recall the slope formula. For the line through $$P(x_1,\,y_1)$$ and $$Q(x_2,\,y_2)$$ the slope is

$$m_{PQ}=\dfrac{y_2-y_1}{x_2-x_1}.$$

Thus

$$m_{PQ}=\dfrac{3-4}{k-1}=\dfrac{-1}{k-1}=-\,\dfrac{1}{k-1}.$$

Two lines are perpendicular if the product of their slopes is $$-1$$. Equivalently, the slope of the perpendicular bisector is the negative reciprocal of $$m_{PQ}$$. Hence

$$m_{\text{perp}}=-(1/m_{PQ})=-(k-1)=-1\cdot\Bigl(-\,\dfrac{k-1}{1}\Bigr)\quad\Longrightarrow\quad m_{\text{perp}}=k-1.$$

Now we write the point-slope form for the perpendicular bisector, using point $$M\Bigl(\dfrac{1+k}{2},\,\dfrac{7}{2}\Bigr)$$:

$$y-\dfrac{7}{2}=(k-1)\Bigl(x-\dfrac{1+k}{2}\Bigr).$$

The problem states that the perpendicular bisector has a $$y$$-intercept of $$-4$$. This means the point $$(0,\,-4)$$ lies on the line. Substituting $$x=0,\;y=-4$$ into the equation of the perpendicular bisector gives

$$-4-\dfrac{7}{2}=(k-1)\Bigl(0-\dfrac{1+k}{2}\Bigr).$$

We simplify the left side:

$$-4=\frac{-8}{2}\quad\Longrightarrow\quad -4-\frac{7}{2}=\frac{-8}{2}-\frac{7}{2}=\frac{-15}{2}.$$

Hence

$$\frac{-15}{2}=(k-1)\Bigl(-\,\dfrac{1+k}{2}\Bigr).$$

Notice the right side contains a product with a negative sign. We can write

$$(k-1)\Bigl(-\,\dfrac{1+k}{2}\Bigr)=-\,\dfrac{(k-1)(1+k)}{2}=-\,\dfrac{k^2-1}{2}.$$

The equation therefore becomes

$$\frac{-15}{2}=-\,\dfrac{k^{2}-1}{2}.$$

Both sides have a common denominator $$2$$, so we multiply the entire equation by $$2$$ to clear the fraction:

$$-15=-(k^{2}-1).$$

Multiplying both sides by $$-1$$ gives

$$15=k^{2}-1.$$

Adding $$1$$ to both sides yields

$$16=k^{2}.$$

Taking the square root of both sides, we obtain two possible values:

$$k=4 \quad\text{or}\quad k=-4.$$

We now compare these with the offered options $$\bigl(-2,\,-4,\,\sqrt{14},\,\sqrt{15}\bigr).$$ Only $$k=-4$$ appears among the choices.

Hence, the correct answer is Option B.