Given a square $ABCD$ with side length $1$, points $P$ and $Q$ are on sides $AB$ and $AD$ respectively. We are to find the angle $\angle PCQ$ given that the perimeter of $\triangle APQ$ is $2$.

Let $AP = x$ and $AQ = y$. Then, $PB = 1 - x$ and $QD = 1 - y$. We need to find $\tan \angle PCQ$.

First, note that:
$$\tan \angle PCQ = \cot(\angle PCB + \angle QCD) = \frac{1 - \tan \angle PCB \tan \angle QCD}{\tan \angle PCB + \tan \angle QCD}.$$

Since $\tan \angle PCB = 1 - x$ and $\tan \angle QCD = 1 - y$, we have:
$$\tan \angle PCQ = \frac{1 - (1 - x)(1 - y)}{(1 - x) + (1 - y)} = \frac{x + y - xy}{2 - x - y}.$$

Given the perimeter condition $x + y + \sqrt{x^2 + y^2} = 2$, we can solve for $\sqrt{x^2 + y^2}$:
$$\sqrt{x^2 + y^2} = 2 - (x + y).$$

Squaring both sides, we get:
$$x^2 + y^2 = (2 - (x + y))^2 = 4 - 4(x + y) + (x + y)^2.$$

Simplifying, we find:
$$x^2 + y^2 = 4 - 4(x + y) + x^2 + 2xy + y^2,$$
$$0 = 4 - 4(x + y) + 2xy,$$
$$2(x + y) = 2 + 2xy,$$
$$x + y - xy = 1.$$

Thus:
$$\tan \angle PCQ = \frac{x + y - xy}{2 - x - y} = \frac{1}{1} = 1,$$
$$\angle PCQ = 45^\circ.$$

The answer is: \boxed{45^\circ}.