We work on the cartesian coordinates. Let $A(-1, 1), B(1, 1), C(1, -1), D(-1, -1), P(x,y)$ be the coordinates. Note that the line $x=y$ is the perpendicular bisector of $AC$, so comparing $PA$ and $PC$ is equivalent to comparing the $x$ and $y$ coordinates of $P$ (i.e. if $x < y$ then $PA < PC$ and if $x > y$ then $PA > PC$. Now WLOG $P$ is closer to $A$ than $C$, so $x \le y$ and the condition now reads as follows: \[PC = \frac{1}{2}(PB + PD) \implies 2\sqrt{(x-1)^2 + (y + 1)^2} = \sqrt{(x-1)^2 + (y - 1)^2} + \sqrt{(x+1)^2 + (y + 1)^2}, \]which after expanding gives us \[4(x^2 + y^2 + 2 + 2(y-x)) = 2(x^2 + y^2 + 2) + 2\sqrt{(x^2 + y^2 + 2 - 2(y+x))(x^2 + y^2 + 2 + 2(y + x))} \]Letting $x^2 + y^2 + 2 = a > 0$, $(y-x) = b \ge 0$, and $x+y = c$ gives us: \[4(a+2b)=2(a) + 2\sqrt{(a-2c)(a+2c)} \implies (a+4b)^2 = a^2 - 4c^2 \le a^2 \le (a+4b)^2. \]Equality must happen, so we must have $c=0$ and $b=0$, which means $(x,y) = (0,0)$ (i.e. the center of $ABCD$) is the only point satisfying the condition.