(TRUE) The statement is True.
$1$. $A$ point on the $y$-axis has coordinates $(0, y)$. Since $P (0, -7)$ lies on the $y$-axis,it satisfies the first condition.
$2$. For $P$ to be on the perpendicular bisector of $AB$,it must be equidistant from $A (-1, 0)$ and $B (7, -6)$.
$3$. Calculate the distance $PA = \sqrt{(0 - (-1))^2 + (-7 - 0)^2} = \sqrt{1^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50}$ units.
$4$. Calculate the distance $PB = \sqrt{(0 - 7)^2 + (-7 - (-6))^2} = \sqrt{(-7)^2 + (-1)^2} = \sqrt{49 + 1} = \sqrt{50}$ units.
$5$. Since $PA = PB$,point $P$ lies on the perpendicular bisector of $AB$. Thus,the statement is true.