Electronic – Should I use a resistor between an input pin of MCU/CPLD and VCC/GND

The input is high-impedance and as such hardly draws any current. But let's, for sake of argument, pretend there flows a (rather large) current of 1\$\mu\$A. This current will flow through the 10k\$\Omega\$ pull-up resistor causing a 10mV (1\$\mu\$A \$\times\$ 10k\$\Omega\$) voltage drop across it. So in this case the voltage on the input pin will be \$V_{CC}\$ - 10mV, probably 5V - 10mV = 4.99V. That will be still recognized as a high level, so no problems here.
The 10k\$\Omega\$ is a typical value for pull-up resistors for this reason: even if there's a small leakage current the voltage drop is negligible. Don't be tempted to increase it to 1M\$\Omega\$, though it will decrease the current when the switch is closed. At 1\$\mu\$A leakage current the voltage drop will be 1\$\mu\$A \$\times\$ 1M\$\Omega\$ = 1V, and then the 5V will drop to 4V. For a 5V supply this will still be OK, but for a 3.3V supply the resulting 2.3V may be too low to be always seen as a high level.

For the pull-down the story is about the same. There doesn't flow any current in the input; you can't say that it would be connected to ground (in which case closing the switch would indeed cause a short-circuit). As such the input takes the voltage you apply to it. If the switch is closed this is \$V_{CC}\$. If the switch is open it's ground (through the pull-down resistor). If there's no current flowing (ideal world) then there's no voltage drop across the resistor either, and the input will be at \$GND\$ level. In a real world situation it may be a few mV.

1. So I'm not too sure I follow this first question here, but I'll take a stab at it: if Vin = 3.3V (i.e. your P-channel mosfet does not connect source and drain) then "TRIGGER" will essentially be disconnected, yes. This assumes that "TRIGGER" is connected to a high-impedance source, like an input pin on a standard microcontroller. As a quick note "essentially disconnected" means that there is a high impedance threshold to get across the source-drain junction, but it is not infinite. If Vin = 0V (i.e. source and drain are connected), then trigger will be close to (but slightly less than) 3.3V

2. So an input pin on a PIC24F is high impedance. This means that it will behave like a resistor with around 1MΩ of resistance to ground. If you check the "Electrical Characteristics" section in the "DC Characteristics: I/O Pin Input Specifications" subsection in the datasheet for the chip you're using, you should find a table which gives you the "Input Leakage Current." Using the dsPIC33FJ256GP710A as an example (it was a datasheet I happened to have on my machine), it lists 0 to 3.5 μA as the leakage current into or out of a pin within the limits of those pins.

   So as a note, you'll also see that in the "Electrical Characteristics" section under the "Absolute Maximum Ratings" subsection, there will be a list which says something to the effect of "Voltage on any pin that is not 5V tolerant with respect to VSS(4) .... -0.3V to (VDD + 0.3V)". This is telling you that you can't put more than ~3.6V on any pin without damaging the pin. This doesn't mean that you should put a current-limiting resistor in front of the input pin, this means you should put a voltage divider in front of the input pin to limit the maximum voltage to less than VDD + 0.3V.

   To tie this into question #1: this means that when a digital input pin is connected to a high-impedance output source like Fig. 1, you will see oscillating input values (1 and 0 randomly) based on when that very small drain removes or adds enough electrons to jump from one digital state to another. This is called a floating input

3. So I just answered this question in the previous one, but I should tell you: the "Maximum output current" is not internally limited. So although the PIC24F can "only output 18mA" this means that if you connect an output pin which is high directly to ground, it will output lots more than 18mA before it burns out! The 18mA is the maximum it can safely output without overheating/damaging the chip. This is true for both sourcing and sinking, there is no fuse which will stop you at 18mA, only the magic blue smoke.

   Just to reiterate: you can safely connect any voltage to a digital input pin within its range (-0.3V to VDD+0.3V for the dsPIC33F). It would not blow the pin to connect it directly to +3.3V, nor would it blow the pin to connect directly to ground.

4. Just answered this in #3, but again, it's not the output impedance of the pin which limits this... Or, more accurately, it is the output impedance, but it will limit the currently by melting.

5. And yes, you can safely remove the 1kΩ resistor from Fig. 2 without damaging your PIC.