6. As shown in Figure 1, in rhombus $A B C D$, let $A E \perp B C$ at point $E, \cos B=\frac{4}{5}, E C=2$, and $P$ is a moving point on side $A B$. Then the minimum value of $P E+P C$ is ( ).
(A) $\frac{2 \sqrt{741}}{5}$
(B) $\frac{2 \sqrt{743}}{5}$
(C) $\frac{2 \sqrt{745}}{5}$
(D) $\frac{2 \sqrt{747}}{5}$
6. C.

Let $AB = x$. Then $BE = x - 2$.
In the right triangle $\triangle ABE$,
$$
\cos B = \frac{BE}{AB} = \frac{x-2}{x} = \frac{4}{5} \Rightarrow x = 10.
$$

By the Pythagorean theorem, it is easy to know that $AE = 6$.
As shown in Figure 5, construct the symmetric point $C'$ of point $C$ with respect to $AB$, and connect $C'E$, which intersects segment $AB$ at point $P$. Then
$$
PE + PC = PE + PC' = C'E
$$

is the minimum value.
Let $CC'$ intersect $AB$ at point $F$, and draw a perpendicular from point $E$ to $CC'$, with the foot of the perpendicular being $G$.
Then $CC' = 2CF = 2AE = 12$.
By the given conditions,
Rt $\triangle CEG$ is similar to Rt $\triangle ABE \Rightarrow \frac{CE}{AB} = \frac{EG}{BE} = \frac{CG}{AE}$
$$
\Rightarrow \frac{2}{10} = \frac{EG}{8} = \frac{CG}{6} \Rightarrow EG = \frac{8}{5}, CG = \frac{6}{5}.
$$

By the Pythagorean theorem,
$$
\begin{array}{r}
C'E = \sqrt{EG^2 + (CC' - CG)^2} \\
= \sqrt{\left(\frac{8}{5}\right)^2 + \left(12 - \frac{6}{5}\right)^2} = \frac{2 \sqrt{745}}{5}.
\end{array}
$$