No, a 5 V zener in that circuit won't work in front of the 5 V regulator. As others have pointed out already, regulators have a dropout voltage, which is the minimum input voltage at which it can maintain its rated output voltage. Usually you think of it in terms of the headroom the regulator needs. The ancient but robust 78xx regulator series is notorious for requiring a lot of headroom, usually around 2.5 V.
However, the zener is a bad idea for another reason. The resistor and zener basically form a shunt regulator. These are very inefficient in that the shunt causes the output voltage to go low by drawing current. A shunt regulator has a constant input current, regardless of what the load is doing. If the load isn't taking the current, then the shunt is.
Power efficiency may not be a big issue, but getting rid of the heat usually is. That is usually bulky and expensive.
The best answer power-wise it to replace the whole thing with a switcher. There are plenty of switcher chips out there that can supply 150 mA from 36 V in, although they aren't quite as simple as a linear regulator. You will have to add a inductor and a few properly selected capacitors. This is all in the datasheet for the switcher chip though, so mostly it's about following the directions carefully.
If you really need the clean output of linear regulator, use a switcher in front of the regulator. It should make a volt or so more than the minimum the linear regulator needs. That gives you some room to put a ferrite chip inductor and a ceramic cap to ground in front of the linear regulator. The active electronics in the linear regulator is good at rejecting noise on the input and not letting it get to the output, but only up to some frequency. The purpose of the little bit of filtering is to eliminate the high frequencies in the switcher output, which then lets the linear regulator deal effectively with the remaining frequencies.
If you really really don't want to use a switcher for some reason (there aren't many good reasons nowadays, ususally this is a religious thing), at least use the zener to make a quick and dirty linear regulator. Something like this:
The zener and R1 still act as a shunt regulator, but the current it draws is so small to be irrelevant with respect to the 150 mA of your load. Q1 and Q2 provide a lot of current gain, probably around 1000 minimum together. That means the R1-D1 shunt regulator only needs to supply 150 µA for the full 150 mA out.
Note that Q2 does the heavy lifting, which is why it is a power transistor. This is still a linear regulator, so you can't escape the fact that (36 V - 5 V) * 150 mA = 4.7 W of heat will end up being dissipated somewhere. By making Q2 drop most of this voltage, it will take most of the heat. I show Q2 as a TIP41, which is a power transistor that comes in a TO-220 package. It can handle that power if provided with a proper heatsink.
Again, a switcher is a better answer here. It will be smaler and cheaper than just heat sink alone.