Let the circumsphere $S$ have center $O$ and radius $R.$ Then we have $\frac{PA}{PA_1}+\frac{PB}{PB_1}+\frac{PC}{PC_1}+\frac{PD}{PD_1}=\frac{PA^2+PB^2+PC^2+PD^2}{p(P,S)}=4$ $\Longrightarrow PA^2+PB^2+PC^2+PD^2=4(R^2-PO^2) \ \ (1)$ Let $G$ be the centroid of $ABCD$ and $a,b,c,d,e,f$ denote its edges. By Leibniz theorem for $P,G$ and $O,G$ we get $PA^2+PB^2+PC^2+PD^2=4PG^2+\frac{_1}{^4}(a^2+b^2+c^2+d^2+e^2+f^2) \ \ (2)$ $OA^2+OB^2+OC^2+OD^2=4OG^2+\frac{_1}{^4}(a^2+b^2+c^2+d^2+e^2+f^2) \ \ (3)$ From $(2)$ and $(3)$ we get $PA^2+PB^2+PC^2+PD^2=4(PG^2+R^2-OG^2) \ \ (4)$ Combining $(1)$ and $(4)$ yields: $4(PG^2+R^2-OG^2)=4(R^2-PO^2)$ $\Longrightarrow PG^2+PO^2=OG^2 \Longrightarrow \angle OPG=90^{\circ}$ Therefore, locus of $P$ is the spherical surface with diameter $\overline{OG}.$

Rewrite the equality as \[ \sum_{\text{cyc}} AP^2 = -4\text{Pow}(P, S) = 4(R^2-OP^2). \]By Leibniz's theorem, we find that \[ \sum_{\text{cyc}} AP^2 = 4R^2+4(PG^2-OG^2), \]where $G$ is the centroid of $ABCD$. Thus, $OG^2+OP^2=PG^2$. Since the prior steps are reversible, the desired locus is the set of points on the sphere with diameter $\overline{PG}$.

By squaring the numerators, the condition translates to $AP^2 + BP^2 + CP^2 + DP^2 = 4(Pow_S(P)) = 4(R^2 - OP^2)$. Also let $s$ be the sum of the squares of the sides of $ABCD$, divided by $4$. So then from Leibniz's theorem we get that $AP^2 + BP^2 + \dots = 4PG^2 + s$ and similarly $4R^2 - 4OG^2 = s$, So then it follows that $4PG^2 + 4R^2 = 4R^2 + 4OG^2 - 4OP^2$, so $PG^2 + PO^2 = OG^2 \implies \angle GPO = 90^\circ$ by Pythagorean. So then the locus of points $P$ is the surface of the sphere with diameter $OG^2$.