You also kinda got your circuit messed up. The (A) point below should be connected to the logic input.
You need a resistor to +5, (A) then the transistor and then ground. Point A, is where the logic point would be,
The issues would be that 0.6 would be dropped across the transistor, to make 0.6 V for a low.
If you wanted to use two diodes to make an or gate, but I think you want an and gate. You can use a diode to do a "wired-or", but not a AND function. See:
https://en.wikipedia.org/wiki/Wired_logic_connection
An AND function could be made with +5, a resistor pull up, (A) an optocoupler (C1), optocoupler (E1), optocoupler (C2), optocoupler (E2) and ground. Point (A) is the logic input. This would make it active low at aprox 1.2 V if they are both triggered at the same time.
What you might be missing is that a diode drops around (0.6 V) and so does the Collector-Emitter junction will drop (0.6V) aproximately for silicon. Schotkey diodes may have a much lower drop (typically (0.2). The drops are approximate. You have to check the datasheet.
I think when I read the phiget datashhet, a low has to less than 1.2 V. If, so some of these methods could be iffy. Two transistors in series add up to 1.2 V.
So, if you want to use one port and AND the result, you might have to use a AND gate.