Let P(h,k) be the point that satisfies the given property. We first consider the perpendicular lines. Two most simple perpendicular lines are

$y=0$ (that is, $x$-axis) and $x=0$ ($y$-axis).

Recall that the distance between a point $(h,k)$ and a line $ax+by+c=0$ is given by

$$ D = \left|\dfrac{ah+bk+c}{\sqrt{a^2+b^2}}\right|$$

Now the distance between the point $P$ and $x$-axis is $|k|$.

The distance between the point $P$ and $y$-axis is $|h|$.

It is given the sum of the above two distances is one. That means,

$$|h|+|k|=1$$

Note that the above equation is a set of four equations, which are

- $h+k=1$, when $P(h,k)$ lies on the first quadrant.
- $-h+k=1$, when $P(h,k)$ lies on the second quadrant.
- $-h-k=1$, when $P(h,k)$ lies on the third quadrant.
- $h-k=1$, when $P(h,k)$ lies on the fourth quadrant.

Note that above four equation represents together a square with corners at $(\pm1,0)$, $(0,\pm 1)$.

The correct answer is $A$.