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

As per the question, we have

$A P=B P$

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

$\Rightarrow \sqrt{(4)^{2}+(2-k)^{2}}=\sqrt{(2+2 k)^{2}+(5)^{2}}$

$\Rightarrow 16+4+k^{2}-4 k=4+4 k^{2}+8 k+25 \quad$ (Squaring both sides)

$\Rightarrow k^{2}+4 k+3=0$

$\Rightarrow(k+1)(k+3)=0$

$\Rightarrow k=-3,-1$

Now for $k=-1$

$A P=\sqrt{(2+2)^{2}+(2-k)^{2}}$

$=\sqrt{(4)^{2}+(2+1)^{2}}$

$=\sqrt{16+9}=5$ units

For $k=-3$

$A P=\sqrt{(2+2)^{2}+(2-k)^{2}}$

$=\sqrt{(4)^{2}+(2+3)^{2}}$

$=\sqrt{16+25}=\sqrt{41}$ units

Hence, $k=-1,-3 ; A P=5$ units for $k=-1$ and $A P=\sqrt{41}$ units for $k=-3$.