1. Given Conditions and Setup:
- We are given a triangle $ABC$ such that $BC = AC + \frac{1}{2} AB$.
- Point $P$ is on $AB$ such that $AP = 3PB$.
- We need to show that $\widehat{PAC} = 2 \widehat{CPA}$.

2. **Construct Points $Q$ and $R$:**
- Let $Q$ be a point on $AC$ such that $AQ = AB$.
- Let $R$ be a point on $BC$ such that $BC \cdot CR = AC \cdot CQ$.

3. Using Length Conditions:
- From the given condition $BC = AC + \frac{1}{2} AB$, we can express the lengths in terms of $AB$ and $AC$.
- Since $AP = 3PB$, let $PB = x$ and $AP = 3x$. Thus, $AB = 4x$.

4. Angle Relationships:
- By construction, $\angle AQB = \frac{1}{2} \angle BAC$ because $AQ = AB$.
- Since $BC \cdot CR = AC \cdot CQ$, we have a similar triangle relationship which implies $\angle ARC = \angle AQB$.

5. Using Cyclic Quadrilateral:
- By the Power of a Point theorem, $BP \cdot BA = BR \cdot BC$.
- This implies that $\angle APC = \angle ARC$.

6. Combining Angles:
- Since $\angle AQB = \frac{1}{2} \angle BAC$ and $\angle ARC = \angle AQB$, we have $\angle APC = \frac{1}{2} \angle BAC$.
- Therefore, $\angle PAC = 2 \angle CPA$.

$\blacksquare$