6. As shown in Figure 1, in the right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, $CA=4$, $P$ is the midpoint of the semicircular arc $\overparen{AC}$, connect $BP$, the line segment $BP$ divides the figure $APCB$ into two parts. The absolute value of the difference in the areas of these two parts is $\qquad$.
Ni.6.4.
As shown in Figure 8, let $A C$ and $B P$ intersect at point $D$, and the point symmetric to $D$ with respect to the circle center $O$ is denoted as $E$. The line segment $B P$ divides the figure $A P C B$ into two parts, and the absolute value of the difference in the areas of these two parts is the area of $\triangle B E P$, which is twice the area of $\triangle B O P$. Thus,
$$
\begin{array}{l}
S_{\triangle B P O}=\frac{1}{2} P O \cdot C O \\
=\frac{1}{2} \times 2 \times 2=2 .
\end{array}
$$

Therefore, the absolute value of the difference in the areas of these two parts is 4.