Let $A, B, C$ be unique collinear points$ AB = BC =\frac13$. Let $P$ be a point that lies on the circle centered at $B$ with radius $\frac13$ and the circle centered at $C$ with radius $\frac13$ . Find the measure of angle $\angle PAC$ in degrees.
1. **Identify the given points and distances:**
   - Points \( A, B, C \) are collinear with \( AB = BC = \frac{1}{3} \).
   - Point \( P \) lies on both the circle centered at \( B \) with radius \( \frac{1}{3} \) and the circle centered at \( C \) with radius \( \frac{1}{3} \).

2. **Determine the position of point \( P \):**
   - Since \( P \) lies on both circles, \( P \) is equidistant from \( B \) and \( C \). Therefore, \( PB = PC = \frac{1}{3} \).

3. **Form the triangle \( PBC \):**
   - In triangle \( PBC \), we have \( PB = PC = BC = \frac{1}{3} \).
   - This implies that triangle \( PBC \) is an equilateral triangle because all sides are equal.

4. **Calculate the angle \( \angle PCB \):**
   - In an equilateral triangle, all internal angles are \( 60^\circ \).
   - Therefore, \( \angle PCB = 60^\circ \).

5. **Use the inscribed angle theorem:**
   - The inscribed angle theorem states that the measure of an angle subtended by an arc at the circumference of a circle is half the measure of the central angle subtended by the same arc.
   - Here, \( \angle PAC \) is an inscribed angle that subtends the same arc as \( \angle PCB \).

6. **Calculate \( \angle PAC \):**
   - Since \( \angle PCB = 60^\circ \), the measure of \( \angle PAC \) is half of \( \angle PCB \).
   - Therefore, \( \angle PAC = \frac{60^\circ}{2} = 30^\circ \).

The final answer is \( \boxed{30^\circ} \).