Let $P$ be an arbitrary point inside $\triangle ABC$ with sides $3,7,8$. What is the probability that the distance of $P$ to at least one vertices of the triangle is less than $1$?

$ \textbf{(A)}\ \frac{\pi}{36}\sqrt 2
\qquad\textbf{(B)}\ \frac{\pi}{36}\sqrt 3
\qquad\textbf{(C)}\ \frac{\pi}{36}
\qquad\textbf{(D)}\ \frac12
\qquad\textbf{(E)}\ \frac 34
$
1. **Calculate the area of the triangle using Heron's formula:**
   - The sides of the triangle are \(a = 3\), \(b = 7\), and \(c = 8\).
   - First, compute the semi-perimeter \(s\):
     \[
     s = \frac{a + b + c}{2} = \frac{3 + 7 + 8}{2} = 9
     \]
   - Using Heron's formula, the area \(A\) of the triangle is:
     \[
     A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{9(9-3)(9-7)(9-8)} = \sqrt{9 \cdot 6 \cdot 2 \cdot 1} = \sqrt{108} = 6\sqrt{3}
     \]

2. **Determine the area of the sectors formed by circles of radius 1 centered at each vertex:**
   - The area of a circle with radius 1 is \(\pi \cdot 1^2 = \pi\).
   - Since the point \(P\) must be within 1 unit of at least one vertex, we consider the sectors formed by these circles.
   - Each sector has an angle of \(60^\circ\) (since the sum of angles in a triangle is \(180^\circ\) and there are three vertices).
   - The area of one sector is:
     \[
     \text{Area of one sector} = \frac{60^\circ}{360^\circ} \cdot \pi = \frac{1}{6} \pi
     \]
   - Since there are three such sectors, the total area of the sectors is:
     \[
     3 \cdot \frac{1}{6} \pi = \frac{1}{2} \pi
     \]

3. **Calculate the probability:**
   - The probability that the distance of \(P\) to at least one vertex is less than 1 is the ratio of the total area of the sectors to the area of the triangle.
   - Therefore, the probability is:
     \[
     \frac{\frac{1}{2} \pi}{6\sqrt{3}} = \frac{\pi}{12\sqrt{3}} = \frac{\pi \sqrt{3}}{36}
     \]

The final answer is \(\boxed{\frac{\pi}{36}\sqrt{3}}\)