Using CMOS logic ICs to drive a P-channel FET
You have a PFET at the output instead of a PNP, so you should not worry about any current going into the PFET's gate. Sure, there is a current (i.e gate charge) that flows into it when turning it on or off and that mainly determines how fast you can operate the PFET.

Now, in the configuration you show, the PFET will always be ON. That is the case because at the output of the CD4081, you can either have a '1' (~12V) or a '0' (0V). The PFET turns on when \$V_S\$ is at a sufficiently high potential with respect to \$V_G\$. Say the output of your CD4081 is at 12V, then \$V_G=6V\$ and \$V_S=12V\$, so \$V_{GS}=-6V\$, that is good enough to turn on lots of PFETs. If on the other hand, the output of your CD4081 is at 0V, then \$V_G=0V\$ and \$V_S=12V\$, so \$V_{GS}=-12V\$, again, that is good enough to turn on most PFETs.

You really want to do something like this with your output PFET:







simulate this circuit &ndash; Schematic created using CircuitLab

The control node is the output of the CD4081. When high (12V), \$V_{GS}=0V\$ and the PFET is OFF. If the control node is low (0V), \$V_{GS}=-12V\$, and the PFET is ON, and the load will see 12V across itself.

EDIT:

I looked at the circuit you linked and I see what you did wrong. This is what you really want to do:







simulate this circuit