6. Given that $P, Q, R, S$ are four points inside the tetrahedron $A-BCD$, and $Q, R, S, P$ are the midpoints of $PA, QB, RC, SD$ respectively. If $V_{P-ABC}$ represents the volume of the tetrahedron $P-ABC$, and similarly for the others, then
$$
V_{P-ABC}: V_{P-BCD}: V_{P-CDA}: V_{P-DAB}=
$$
$\qquad$
$6.8: 1: 2: 4$.
As shown in Figure 5, let the distances from points $A, P$ to the plane $BCD$ be $h_{1}, h_{P}$, and similarly for the others. Thus, we have
$$
\begin{array}{c}
h_{Q}=\frac{1}{2}\left(h_{A}+h_{P}\right), \\
h_{R}=\frac{1}{2} h_{Q} \\
=\frac{1}{4}\left(h_{1}+h_{P}\right), \\
h_{S}=\frac{1}{2} h_{R}=\frac{1}{8}\left(h_{A}+h_{P}\right), \\
h_{P}=\frac{1}{2} h_{S}=\frac{1}{16}\left(h_{A}+h_{P}\right),
\end{array}
$$

Therefore, $h_{P}=\frac{1}{15} h_{1}$.
Hence, $V_{P-BCD}: V_{A-BCD}=1: 15$.
Similarly, $V_{P-CDA}: V_{A-BCD}=2: 1$,
$$
V_{P-DAB}: V_{A-BCD}=4: 15, V_{P-ABC}: V_{A-BCD}=8: 15 \text {. }
$$

Therefore, the required ratio is $8: 1: 2: 4$.