Given an equilateral triangle with side length $10$ and a point $P$ inside the triangle, we are required to find the distance from $P$ to the third side, knowing the distances from $P$ to the other two sides are $1$ and $3$.

First, recall the area formula of a triangle in terms of its base and corresponding height:

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}.$$

The area of an equilateral triangle with side length $s$ is given by:

$$\text{Area} = \frac{\sqrt{3}}{4} s^2.$$

For our equilateral triangle with side length $10$:

$$\text{Area} = \frac{\sqrt{3}}{4} \times 10^2 = 25\sqrt{3}.$$

Consider point $P$ to have distances $d_1 = 1$ and $d_2 = 3$ from two sides of the triangle. Let $d_3$ be the distance from $P$ to the third side. The area of the triangle can also be expressed as the sum of the areas formed by dropping perpendiculars from $P$ to each of the three sides. Thus:

$$25\sqrt{3} = \frac{1}{2} \times 10 \times 1 + \frac{1}{2} \times 10 \times 3 + \frac{1}{2} \times 10 \times d_3.$$

Simplifying, we have:

$$25\sqrt{3} = 5(1 + 3 + d_3).$$

$$25\sqrt{3} = 5(4 + d_3).$$

Divide both sides by 5:

$$5\sqrt{3} = 4 + d_3.$$

Solving for $d_3$, we get:

$$d_3 = 5\sqrt{3} - 4.$$

Therefore, the distance from $P$ to the third side is:

$$\boxed{5\sqrt{3} - 4}.$$