In a parallel plate capacitor, how does placing a metal in between the plates reduce the capacitance?

In the case of the valve/tube, the purpose of the screen grid is to reduce the effect of the capacitance between the grid and anode.

The screen grid is not just a floating piece of metal, it's connected to a low impedance supply (don't remember offhand whether it's low or high voltage).

Without it, when the anode changes voltage, the anode to grid capacitance induces a current into the grid which fights the control signal. This is the so called Miller capacitance effect. This loads the grid with an effective capacitance that's enhanced by the gain of the valve.

With the screen grid, the anode to screen capacitance induces that current into the screen grid. Being supplied with a low impedance, that current flows into the supply with little change in voltage. The screen to grid capacitor sees little change in voltage, and so the grid has much, much less current induced it, possibly being only enhanced by a very small gain, often less than 2.

If the screen grid is left unconnected, then sure, anode to grid capacitance remains essentially unaltered, as does the Miller effect when the valve is used as an amplifier.

The capacity remains the same, as seen by the calculation below.

$$ C1+C2=\dfrac{\dfrac{ϵA}{d1} \cdot \dfrac{ϵA}{d2}}{\dfrac{ϵA}{d1}+ \dfrac{ϵA}{d2}}=\dfrac{ϵA}{d1+d2}=C $$

You are forgetting one factor about elements in series, the voltage is also split between the capacitors while the charge stays the same, imagine you got capacitor A and capacitor B

$$ V_{total}= V_a+V_b=\frac{Q_T}{C_A}+\frac{Q_T}{C_B} $$

if we rearrange $$ \frac{V_{total}}{Qt}=\frac{1}{C_T}=\frac{1}{C_A}+\frac{1}{C_B} $$

which ends up looking like resistors in parallel operation wise. If they are identical you get half.