If the point P(k − 1, 2) is equidistant from the points A(3, k) and B(k, 5),

The given points are P(k − 1, 2), A(3, k) and B(k, 5).

$\because A P=B P$

$\therefore A P^{2}=B P^{2}$

$\Rightarrow(k-1-3)^{2}+(2-k)^{2}=(k-1-k)^{2}+(2-5)^{2}$

$\Rightarrow(k-4)^{2}+(2-k)^{2}=(-1)^{2}+(-3)^{2}$

$\Rightarrow k^{2}-8 y+16+4+k^{2}-4 k=1+9$

$\Rightarrow k^{2}-6 y+5=0$

$\Rightarrow(k-1)(k-5)=0$

$\Rightarrow k=1$ or $k=5$

Hence, k = 1 or k = 5.