Let $Q$ be the point such that $ABQP$ and $CDPQ$ are parallelograms. Then Ptolemy's inequality implies $$AP \cdot CP + BP \cdot DP = BQ \cdot CP + BP \cdot CQ \geq BC \cdot PQ = 1.$$Equality holds if and only if $B,Q,C,P$ are concyclic in this order. In particular, $P$ must be inside of $ABCD$ and $PX \cdot QX = BX \cdot CX$ where $X= PQ \cap BC$. If $Y$ is the foot of the perpendicular from $P$ to $AB$ then this implies $BX \cdot CX = AY \cdot BY$. Since $BX+CX=AY+BY=1$, this means that either $AY=BX$ or $BY=BX$. The first equation holds iff $P$ lies on $AC$, and the second one holds iff $P$ lies on $BD$. All in all, the locus is the union of the diagonals $AC$ and $BD$.

Let's consider the situation on complex plane. We can take WLOG $a=\sqrt2$ and $$A=-1,\ B=-i,\ C=1,\ D=i.$$Equation is equivalent to $$2=|p+1|\cdot |p-1|+|p+i|\cdot |p-i|\iff 2= |p^2+1|+ |1-p^2|.$$By triangle inequality $$|p^2+1|+ |1-p^2|\ge |p^2+1+1-p^2|=2$$with equality iff $$p^2+1=0\vee p^2-1=0\vee \exists_{t\in\mathbb{R}}\ t(p^2+1)=1-p^2$$Now observe $$p^2+1=0\iff (p=i\vee p=-i)\iff P\in \lbrace B,D\rbrace$$$$p^2-1=0\iff (p=1\vee p=-1)\iff P\in \lbrace A,C\rbrace.$$And we're left with the last condition. $$\exists_{t\in\mathbb{R}_+}\ t(p^2+1)=1-p^2\iff \exists_{t\in\mathbb{R}_+}\ p^2=\frac{1-t}{1+t}\iff p^2\in(-1,1)\iff$$$$ \exists_{q\in(-1;1)}\left(p=q\cdot i\vee p=q\right)\iff\left( (P\in \overline{AC}\wedge P\neq A\wedge P\neq C)\vee (P\in \overline{BD}\wedge P\neq B\wedge P\neq D)\right)$$QED

As you can see this isn't full answer. Gryphos wrote: All in all, the locus is the union of the diagonals $AC$ and $BD$.