1. **Express $x$ and $y$ in polar form:**
Let $x = ae^{is}$ and $y = be^{it}$, where $a = |x|$ and $b = |y|$.

2. Calculate the modulus of the given expression:
$$P = \left|\frac{x + y}{1 + x\overline{y}}\right|$$
Since $\overline{y} = be^{-it}$, we have:
$$P = \left|\frac{ae^{is} + be^{it}}{1 + abe^{i(s-t)}}\right|$$

3. Simplify the numerator and the denominator:
$$P^2 = \left|\frac{ae^{is} + be^{it}}{1 + abe^{i(s-t)}}\right|^2 = \frac{(ae^{is} + be^{it})(ae^{-is} + be^{-it})}{(1 + abe^{i(s-t)})(1 + abe^{-i(s-t)})}$$

4. Expand the numerator and the denominator:
$$(ae^{is} + be^{it})(ae^{-is} + be^{-it}) = a^2 + b^2 + ab(e^{i(s-t)} + e^{-i(s-t)})$$
$$(1 + abe^{i(s-t)})(1 + abe^{-i(s-t)}) = 1 + a^2b^2 + ab(e^{i(s-t)} + e^{-i(s-t)})$$

5. Introduce a new variable for simplification:
Let $u = e^{i(s-t)} + e^{-i(s-t)} = 2\cos(s-t)$, where $u \in [-2, 2]$.

6. **Rewrite the expression in terms of $u$:**
$$P^2 = \frac{a^2 + b^2 + abu}{1 + a^2b^2 + abu}$$

7. **Analyze the function $f(u) = \frac{a^2 + b^2 + abu}{1 + a^2b^2 + abu}$:**
$$P^2 = 1 - \frac{(a^2 - 1)(b^2 - 1)}{1 + a^2b^2 + abu}$$

8. **Determine the conditions for $P$:**
- If $(a^2 - 1)(b^2 - 1) > 0$, then $P_{\text{max}} = \frac{a + b}{1 + ab}$.
- If $(a^2 - 1)(b^2 - 1) < 0$, then $P_{\text{max}} = \frac{|a - b|}{|1 - ab|}$.
- If $(a^2 - 1)(b^2 - 1) = 0$, then $P = 1$.

The final answer is $\boxed{\frac{a + b}{1 + ab}}$ or $\boxed{\frac{|a - b|}{|1 - ab|}}$ or $\boxed{1}$ depending on the conditions.