Points $P$ and $Q$ are chosen on diagonal $AC$ of square $ABCD$ such that $AB=AP=CQ=1$. What is the measure of $\angle{PBQ}$ in degrees?

[i]2019 CCA Math Bonanza Lightning Round #1.3[/i]
1. Given that \(ABCD\) is a square with side length \(AB = 1\), and points \(P\) and \(Q\) are chosen on the diagonal \(AC\) such that \(AP = AB = 1\) and \(CQ = 1\).
2. Since \(ABCD\) is a square, the diagonal \(AC\) can be calculated using the Pythagorean theorem:
   \[
   AC = \sqrt{AB^2 + BC^2} = \sqrt{1^2 + 1^2} = \sqrt{2}
   \]
3. Point \(P\) is on \(AC\) such that \(AP = 1\). Since \(AP = AB\), triangle \(ABP\) is isosceles with \(AB = AP\).
4. In an isosceles triangle \(ABP\) with \(AB = AP\), the base angles are equal. The angle at \(A\) in square \(ABCD\) is \(90^\circ\), so:
   \[
   \angle ABP = \angle APB = \frac{180^\circ - 90^\circ}{2} = 45^\circ
   \]
5. Similarly, point \(Q\) is on \(AC\) such that \(CQ = 1\). Since \(CQ = AB\), triangle \(BCQ\) is isosceles with \(BC = CQ\).
6. In an isosceles triangle \(BCQ\) with \(BC = CQ\), the base angles are equal. The angle at \(C\) in square \(ABCD\) is \(90^\circ\), so:
   \[
   \angle BCQ = \angle CBQ = \frac{180^\circ - 90^\circ}{2} = 45^\circ
   \]
7. Now, we need to find \(\angle PBQ\). Since \(P\) and \(Q\) lie on the diagonal \(AC\), and \(AB = AP = CQ = 1\), the triangles \(ABP\) and \(BCQ\) are congruent by the Side-Angle-Side (SAS) criterion.
8. Therefore, \(\angle PBQ\) is the angle between the two congruent triangles at point \(B\). Since each of the base angles is \(45^\circ\), the angle between them is:
   \[
   \angle PBQ = 45^\circ
   \]

The final answer is \(\boxed{45^\circ}\).