The perpendicular bisector of the line segment joining $P(1,4)$ and $Q(k,3)$ has $y$ -intercept $-4$. Then a possible value of $k$ is:

Midpoint of $PQ=\left(\frac{k+1}{2},\frac{4+3}{2}\right)=\left(\frac{k+1}{2},\frac{7}{2}\right)$.Slope of $PQ=m_1=\frac{3-4}{k-1}=\frac{-1}{k-1}$.Slope of the perpendicular bisector $m_2=-\frac{1}{m_1}=k-1$.Equation of the perpendicular bisector:
$$y-\frac{7}{2}=(k-1)\left(x-\frac{k+1}{2}\right)$$
The $y$ -intercept is found by putting $x=0$:
$$-4-\frac{7}{2}=(k-1)\left(-\frac{k+1}{2}\right)$$
$$-\frac{15}{2}=-\frac{k^2-1}{2}$$
$$15=k^2-1$$
$$k^2=16⟹k=\pm 4$$
So, a possible value is $-4$.