Answer
We show that $\nabla {f_A}\left( P \right) = {{\bf{e}}_{AP}}$.
Work Step by Step
We are given a fixed point $A = \left( {a,b} \right)$ in the plane and there is a point $P = \left( {x,y} \right)$. Write the distance from $A$ to $P$:
${f_A}\left( P \right) = \sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}} $
Let ${{\bf{e}}_{AP}}$ be the unit vector pointing from $A$ to $P$. So,
${{\bf{e}}_{AP}} = \frac{{\overrightarrow {AP} }}{{||\overrightarrow {AP} ||}} = \frac{{\left( {x,y} \right) - \left( {a,b} \right)}}{{||\left( {x,y} \right) - \left( {a,b} \right)||}} = \frac{{\left( {x - a,y - b} \right)}}{{\sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}} }}$
${{\bf{e}}_{AP}} = \left( {\frac{{x - a}}{{\sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}} }},\frac{{y - b}}{{\sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}} }}} \right)$
Next, we evaluate the gradient of ${f_A}\left( P \right)$:
$\nabla {f_A}\left( P \right) = \left( {\frac{{\partial {f_A}\left( P \right)}}{{\partial x}},\frac{{\partial {f_A}\left( P \right)}}{{\partial y}}} \right)$
$\nabla {f_A}\left( P \right) = \left( {\frac{{x - a}}{{\sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}} }},\frac{{y - b}}{{\sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}} }}} \right)$
Thus, $\nabla {f_A}\left( P \right) = {{\bf{e}}_{AP}}$.
