Given points

$P(1,4), \quad Q(k,3)$

We need the perpendicular bisector of $PQ$, and it is given that its y-intercept is $-4$.

Midpoint of $PQ$

The perpendicular bisector passes through the midpoint of $PQ$.

Midpoint: $M\left(\frac{1+k}{2},\frac{4+3}{2}\right)=\left(\frac{1+k}{2},\frac{7}{2}\right)$

Slope of $PQ$

$m_{PQ}=\frac{3-4}{k-1}=\frac{-1}{k-1}$

So the slope of the perpendicular bisector is the negative reciprocal: $m_\perp = k-1$

Equation of the perpendicular bisector

Using point-slope form through midpoint $M$: $y-\frac{7}{2}=(k-1)\left(x-\frac{1+k}{2}\right)$

Use the y-intercept condition

At the y-intercept, $x=0$ and $y=-4$.

Substitute into the equation: $-4-\frac{7}{2}=(k-1)\left(0-\frac{1+k}{2}\right)$

$-\frac{15}{2}=-(k-1)\frac{k+1}{2}$

Multiply by $2$: $-15=-(k-1)(k+1)$

$15=k^2-1$

$k^2=16$

$k=\pm 4$

Check the options

Options are:

• A: $1$
• B: $2$
• C: $-2$
• D: $-4$

Possible values are $k=4$ or $k=-4$. Among the given options, only $-4$ is present.

Conclusion

The correct option is: $\boxed{-4}$