We are asked to find the value of the angle marked $P$ in the diagram. The diagram shows an equilateral triangle with one vertex cut off by a line segment. We are given that the angle between the line segment and one side of the triangle is $130^{\circ}$.

1. Problem Description

We are asked to find the value of the angle marked $P$ in the diagram. The diagram shows an equilateral triangle with one vertex cut off by a line segment. We are given that the angle between the line segment and one side of the triangle is $130^{\circ}$.

2. Solution Steps

First, notice that the triangle with three sides marked with identical ticks is an equilateral triangle. Therefore, each of the angles in this triangle is $60^{\circ}$.

Let the equilateral triangle be $ABC$. Let the line segment intersect $AC$ at $D$ and $BC$ at $E$. We are given that $\angle ADE = 130^{\circ}$ and we need to find $\angle DEC = P$.

Since $\angle ADE$ and $\angle CDE$ form a straight line, their sum is $180^{\circ}$. Therefore, $\angle CDE = 180^{\circ} - 130^{\circ} = 50^{\circ}$.

Now, consider triangle $CDE$. We know that $\angle DCE = \angle ACB = 60^{\circ}$ and $\angle CDE = 50^{\circ}$. Since the sum of the angles in a triangle is $180^{\circ}$, we have

$$\angle CED = 180^{\circ} - \angle DCE - \angle CDE = 180^{\circ} - 60^{\circ} - 50^{\circ} = 70^{\circ}$$

Since $\angle CED = P$, we have $P = 70^{\circ}$.

3. Final Answer

The value of the angle $P$ is $70^{\circ}$.