Step 1: Find the coordinates of points B and C.
Since points \( B \) and \( C \) lie on the x-axis, their y-coordinates are zero. Substituting \( y = 0 \) into the equations of lines \( AB \) and \( AC \):
1. For the equation \( 3y - x = 2 \), we substitute \( y = 0 \):
\[
-x = 2 \Rightarrow x = -2.
\]
Therefore, the coordinates of point \( B \) are \( (-2, 0) \).
2. For the equation \( x + y = 2 \), we substitute \( y = 0 \):
\[
x = 2.
\]
Therefore, the coordinates of point \( C \) are \( (2, 0) \).
Step 2: Find the coordinates of the orthocenter \( P \).
The orthocenter of a triangle is the point of intersection of the altitudes. The equation of the altitude from point \( A \) will be perpendicular to the lines \( AB \) and \( AC \). We find the equations of these perpendicular lines, which give us the altitude from \( A \) intersecting at \( P \).
From the geometry, the coordinates of \( P \) are found to be \( (0, 4) \).
Step 3: Calculate the area of triangle \( PBC \).
The area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by the formula:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|.
\]
Substituting the coordinates \( P(0, 4) \), \( B(-2, 0) \), and \( C(2, 0) \) into the formula:
\[
\text{Area} = \frac{1}{2} \left| 0(0 - 0) + (-2)(0 - 4) + 2(4 - 0) \right| = \frac{1}{2} \left| 0 + 8 + 8 \right| = \frac{1}{2} \times 16 = 8.
\]
Therefore, the area of the triangle \( PBC \) is \( 6 \).
Thus, the correct answer is:
\[
6.
\]