Yep, this joystick interface looks like a crude Kempston compatible one.
Even though it uses a 4025 triple 3-input logic chip for the address and control decoding, only one of the three 3-input NOR gates is being used. So the partial decoding is only when A5 is low (0), A7 is low (0) and /IORQ is active (low).
Kempson interfaces vary from only needing A7 low (0) to having all 8 bits of address lines A0 to A7 correctly decoded. But ZX Spectrum software is written to use port 31 (dec) 0x1F (hex).
The interface will respond if the CPU tries to write to any I/O port address with A5 and A7 low, as it ignores the Z80 /RD control line. Writing will cause a bus collision as the Z80 and the 4066 chips both try to outdrive each other (read as the 4066 shorting out the CPU) which is poor design
With this design, it looks like pin 8 (common) on the 9-way D-connector is connected to +5V. So don't try to use an autofire joystick. Pin 8 is normaly 0V/GND on Atari machines.
There are some strange things with this interface. First, despite there being five 560 ohm pull down resistors (those on the front of the board), four extra 150 ohm pull down resistors have been soldered to the track side in parallel with four of the 560 ohm resistors. Why add these? And why only four? The down direction input does not have an extra 150 ohm resistor.
In the photos, I presume the black wire has come adrift from pin 8 of the 9 way D-connector.
The tracks for the pins of the 4025 chip look like pins 1 to 7 are all connected together, but pin 6 is an output pin! [pins 1 to 5 are inputs and pin 7 is the GND pin for the chip]. I also can't see the track connecting this group to the 0V/GND rail. But it may just not be visible in the photo.
And last, edge-connector pins 10 and 11 look like they have either got a track between them, or a bit of solder. One is a data line (pin 10), the other is the /INT pin. They should not be connected together.