Hi Jinx,
Due the need for lowest possible current on this system and the half
wave rectifier, your 9V zener could transfer a high ripple to the final
5V zener.  I wonder if the ripple could not reach the limits of 5V
already.

To avoid this possible problem, I would increase the system filtering,
but to do that you need to increase the power drain from the source,
what would need a bigger input cap and lower impedances in general, or,
increase the 9V zener to a little higher voltage one, as 12V for
example.

The use of two zeners is a required safety procedure, since if the first
zener blows up (open), your final circuit still protected by the second
zener.  I used a double configuration like that for a power meter with
0-1mA output, the Vdc is 12V and the primary zener is 33V.

You would want to use a HSF (low ESR) capacitor at the 5V PIC input,
with a couple of 0.1µF.

The input cap (0.22µF) have different charge current for each senoid
polarity. This is caused by your 9V zener that offers a 9V drop in one
side of the senoid, while only 0.6V at the other senoid side. Remember
that even with a nice current consume at the load, it is not the same as
the full direct current flow through the zener. 

I used a 1N4007 diode in series with the zener, with a resistor in
parallel to this diode. This resistor value is to balance both senoid
polarity current over the cap.  Without much calculation, I guess the
primary cap current is around 9mA, so a 820 Ohms resistor should
generate a Vdrop enough to balance when the senoid is negative at the
hot wire.  You need to test it in the bench.

If you don't balance it, the cap will charge more than discharge,
reducing the efficiency.

       |
       |
     ,---'
       A    12Vdc Zener
       |
       o---. 
       |   |
1N     |   R  820 Ohms
4007  _V_  R
       |   |
       o---'
       |
       |