Depletion P-channel JFET that's saturated at \$V_{GS}=0V\$
From your question I suspect you are missing something about JFETs structure or operation. 

JFETs are implicitly depletion-mode devices, i.e. there are no "enhancement" JFETs. At \$V_{GS}=0V\$ any JFET (P- or N-channel) will be heavily conducting, since a channel is already formed (depletion-mode). The only difference between the two polarities is that the channel is "filled" with different type of charges (holes for P-channel, electrons for N-channel).

Here you can see the output characteristics of a common through-hole P-channel JFET (J176):



The top curve corresponds to 0V gate-source voltage. J176 is designed for switching applications, so its characteristics are plotted for relatively low values of \$V_{DS}\$. A P-channel JFET for linear applications is the MMBF5460:



again the top curve is for \$V_{GS}=0V\$.

You can see, especially in the second example, that for sufficiently high values of \$\left|V_{DS}\right|\$ the device enters saturation, i.e. the drain current becomes almost independent from \$V_{DS}\$ (the curve becomes nearly horizontal).

In summary, the device can be saturated if operated with high enough \$V_{DS}\$. In particular the condition to be saturated would be:

\[
\left|V_{DS}\right| > \left|V_{GS} - V_{GS(off)} \right|
\]

where \$V_{GS(off)}\$ is found in the datasheet.