$ABCD$ is a square of side 1. $P$ and $Q$ are points on $AB$ and $BC$ such that $\widehat{PDQ} = 45^{\circ}$. Find the perimeter of $\Delta PBQ$.
1. **Define the problem and variables:**
   - Let $ABCD$ be a square with side length 1.
   - Points $P$ and $Q$ are on sides $AB$ and $BC$ respectively.
   - Let $AP = x$ and $CQ = y$, where $0 \leq x, y \leq 1$.
   - Therefore, $BP = 1 - x$ and $BQ = 1 - y$.
   - We need to prove that the perimeter of $\triangle PBQ$ is constant given that $\angle PDQ = 45^\circ$.

2. **Calculate the area of $\triangle PDQ$:**
   - The area of $\triangle PDQ$ can be expressed using the sine area formula:
     \[
     [PDQ] = \frac{1}{2} \cdot DP \cdot DQ \cdot \sin 45^\circ
     \]
   - Since $\sin 45^\circ = \frac{\sqrt{2}}{2}$, we have:
     \[
     [PDQ] = \frac{1}{2} \cdot DP \cdot DQ \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4} \cdot DP \cdot DQ
     \]

3. **Express the area of $\triangle PDQ$ in terms of $x$ and $y$:**
   - The area of the square $ABCD$ is 1.
   - The area of $\triangle APD$ is $\frac{x}{2}$.
   - The area of $\triangle BQP$ is $\frac{(1-x)(1-y)}{2}$.
   - The area of $\triangle CDQ$ is $\frac{y}{2}$.
   - Therefore, the area of $\triangle PDQ$ is:
     \[
     [PDQ] = 1 - \frac{x}{2} - \frac{(1-x)(1-y)}{2} - \frac{y}{2}
     \]

4. **Simplify the area expression:**
   \[
   [PDQ] = 1 - \frac{x}{2} - \frac{1 - x - y + xy}{2} - \frac{y}{2}
   \]
   \[
   [PDQ] = 1 - \frac{x}{2} - \frac{1}{2} + \frac{x}{2} + \frac{y}{2} - \frac{xy}{2} - \frac{y}{2}
   \]
   \[
   [PDQ] = 1 - \frac{1}{2} - \frac{xy}{2}
   \]
   \[
   [PDQ] = \frac{1}{2} - \frac{xy}{2}
   \]

5. **Equate the two expressions for $[PDQ]$:**
   \[
   \frac{\sqrt{2}}{4} \cdot DP \cdot DQ = \frac{1}{2} - \frac{xy}{2}
   \]
   \[
   DP \cdot DQ \cdot \frac{\sqrt{2}}{2} = 1 - xy
   \]
   \[
   DP \cdot DQ \cdot \cos 45^\circ = 1 - xy
   \]

6. **Apply the generalized Pythagorean theorem to $\triangle PDQ$:**
   \[
   PQ^2 = DP^2 + DQ^2 - 2 \cdot DP \cdot DQ \cdot \cos 45^\circ
   \]
   \[
   PQ^2 = 1 + x^2 + 1 + y^2 - 2(1 - xy)
   \]
   \[
   PQ^2 = 2 + x^2 + y^2 - 2 + 2xy
   \]
   \[
   PQ^2 = (x + y)^2
   \]
   \[
   PQ = x + y
   \]

7. **Calculate the perimeter of $\triangle PBQ$:**
   \[
   \text{Perimeter of } \triangle PBQ = BP + BQ + PQ
   \]
   \[
   = (1 - x) + (1 - y) + (x + y)
   \]
   \[
   = 1 - x + 1 - y + x + y
   \]
   \[
   = 2
   \]

Conclusion:
The perimeter of $\triangle PBQ$ is constant and equal to 2.

The final answer is $\boxed{2}$