6. As shown in Figure 1, in the right triangular prism $A B C-A_{1} B_{1} C_{1}$, $\angle A C B=90^{\circ}, B C=C C_{1}$ $=2, A C=4 \sqrt{2}, P$ is a point on $B C_{1}$. Then $C P+P A_{1}$ 
Figure 1
$6.2 \sqrt{13}$.
From the problem, we know that $\triangle C C_{1} B$ is an isosceles right triangle.
Also, $A_{1} C_{1} \perp$ plane $B C_{1}$, so $\angle A_{1} C_{1} B=90^{\circ}$.
Unfolding the dihedral angle $A_{1}-B C_{1}-C$ along $B C_{1}$ into a plane figure, we get quadrilateral $A_{1} C_{1} C B$, as shown in Figure 2.

Therefore, $C P+P A_{1}$ achieves its minimum value if and only if $C P A_{1}$ is the diagonal of quadrilateral $A_{1} C_{1} C B$.
From the problem, we know
Figure 2
$$
\begin{array}{l}
\angle C C_{1} A=45^{\circ}+90^{\circ} \\
=135^{\circ} .
\end{array}
$$

According to the cosine rule, we have
$$
\begin{array}{l}
A_{1} C^{2}=(4 \sqrt{2})^{2}+2^{2}-2 \times 4 \sqrt{2} \times 2 \cos 135^{\circ} \\
=52 \\
\Rightarrow A_{1} C=2 \sqrt{13} .
\end{array}
$$

Thus, the minimum value of $C P+P A_{1}$ is $2 \sqrt{13}$.