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Why the diode in the circuit?

When the switch opens the falling edge will cause a negative pulse on the 4013 input. The diode removes that negative pulse.

Input to 4013 without diode – one pulse (on/off) – zero volts is the centre line. The signal goes negative to minus 1.44 volts. Positive pulse comes from direct connection to +3 volts, negative pulse is smaller because capacitor has to discharge through 1M resistor.
P7112944

Input to 4013 with the diode – the negative signal is almost totally removed. The signal goes negative to minus 0.35 volts.
P7112945

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