Assume such a point $P$ exists. Let $O = (0, 0)$, $A = (x, 0)$, $B = (x, y)$, and $C = (0, y)$ where $x$ and $y$ are positive odd integers. Let $a, b, c$, and $d$ be the distances from $P$ to $OC$, $AB$, $BC$, and $OA$ respectively. So $x = a + b$ and $y = c + d$ and $P = (a, d)$. Then $PA = \sqrt{a^2 + c^2}$ and $PB = \sqrt{(x - a)^2 + c^2}$, so $a^2 + c^2$ and $a^2 + c^2 - 2ax + x^2$ are perfect squares. Thus, $2ax$ is an integer, which implies $a$ is rational with a denominator not divisible by $4$ (when in lowest terms). Similarly, $b, c$, and $d$ are also rational with denominators not divisible by $4$ (when in lowest terms). If $k$ is the denominator of $a$ and $b$, while $l$ is the denominator of $c$ and $d$, we scale the rectangle through $O$ by $lcm(k, l)$ (note this cannot be divisible by $4$). Let $X'$ denote the point $X$ after scaling and $z'$ denote the distance $z$ after scaling. So $A' = (x', 0)$, $B' = (x', y')$, $C' = (0, y')$, and $P' = (a', d')$, implying that $a', d', c' = y' - d'$, and $b' = x' - a'$ are positive integers. Also, $P'O$, $P'A'$, $P'B'$, and $P'C'$ must be positive integers. Either $x'$ and $y'$ are odd (if $2 \nmid lcm(k, l)$), or both are $2 \pmod{4}$ (if $2 \mid lcm(k, l)$). If $x'$ and $y'$ are both odd, then one of $a'$ or $b'$ and one of $c'$ or $d'$ must be odd. Thus, one of $a'^2 + c'^2$, $a'^2 + d'^2$, $b'^2 + c'^2$, or $b'^2 + d'^2$ must be $2 \pmod{4}$, which cannot be a perfect square, so one of $P'O$, $P'A'$, $P'B'$, or $P'C'$ cannot be an integer, contradiction. If $x'$ and $y'$ are both $2 \pmod{4}$, then at least one of $k$ or $l$ is even (and $2 \pmod{4}$). If both $k$ and $l$ are $2 \pmod{4}$, then since $a, b, c, d$ were multiplied by a number $2 \pmod{4}$, we must have that $a’, b’, c’, d’$ are all odd. Hence, $a'^2 + c'^2$, $a'^2 + d'^2$, $b'^2 + c'^2$, and $b'^2 + d'^2$ are all $2 \pmod{4}$, which cannot be perfect squares, so $P'O$, $P'A'$, $P'B'$, and $P'C'$ cannot be integers, contradiction. If exactly one of $k$ or $l$ is even, WLOG assume $k$ is odd while $l$ is even. So $a’$ and $b’$ must be even while $c’$ and $d’$ must be odd. But $a’ + b’ \equiv 2 \pmod{4}$, so one of $a’$ or $b’$ must be $0 \pmod{4}$, while the other is $2 \pmod{4}$. So one of $a’^2$ or $b’^2$ must be $0 \pmod{8}$, while the other is $4 \pmod{8}$. Meanwhile, $c’^2$ and $d’^2$ are $1 \pmod{8}$. Thus, one of $a'^2 + c'^2$, $a'^2 + d'^2$, $b'^2 + c'^2$, or $b'^2 + d'^2$ must be $5 \pmod{8}$, which cannot be a perfect square, so one of $P'O$, $P'A'$, $P'B'$, or $P'C'$ cannot be an integer, contradiction. Therefore, such a point $P$ does not exist.