You told that your current consumption is up to about 200mA. This makes it difficult to use a NiCd-battery, because only about 1/100 of nominal current is allowed to flow into the cell when being permanently connected to the charger. Means if you have a NiCd-battery with capacity of 1Ah, then only 10mA is allowed to flow. This is normally guaranteed by the use of a current limiting resistor between charger and NiCd-battery. (This resistor is omitted in your design, by the way, which makes your application a bit dangerous.)
NiCd-batteries which are charged by such small currents, and which can keep connected to charger all the time, are only used, when the battery backup must only provide very low currents, for instance, if you have a digital circuit using a RTOS (real time clock oscillator). Then the current is so little, that charging the battery by 1/100 of nominal current is enough to keep the battery always fully charged.
But if you need much higher currents from the backup battery, then the battery must be charged with much higher currents too. In this case a Pb-battery is a much better choice, because it's much simpler to charge a Pb-battery than a NiCd-battery.
In the circuit I recommended you, and which I recall here again


the charging is done by the LM317. End voltage of charging is 13.75V = 1.25V x ((1200 + 120) / 120) (no current is flowing through 4R3 resistor then), so that there's no risk to destroy the Pb-battery. Charging current is limited to about 200mA at 4V output voltage (at anode of diode), to 120mA at 8V, and 30mA at 12V. Current can be calculated as follows:
Uout at output causes a current to flow through 120R and 1k2 resistors of Uout /(1200 Ohm + 120 Ohm). This current causes a voltage drop at 120R resistor of Uout / (1200 Ohm + 120 Ohm) x 120 Ohm. Regulator tries to keep voltage drop across the series of 4R3 and 120R resistors constant to 1.25V. So, the current through 4R3 resistor is (1.25V - Uout / (1200 Ohm + 120 Ohm) x 120 Ohm) / 4.3 Ohm.

Mains transformer must be capable to deliver 400mA, 200mA for the circuited powered by the 9V regulator and 200mA for the charger (worst case). If your mains transformer is strong enough, then you can slightly increase the charging current. This can be helpful, if current is drawn by the battery more than seldom.
If you plan to often power the circuit by the backup battery, on the other hand, then charging current must be much higher than in the circuit above. But then, a more sophisticated methode is needed, which furtherly limits the charging current in the case of fully discharged battery.

To find out, whether a (charged) battery is connected or not, make the Px,z output to emit logic high state. This turns-on the MOSFET BS170, which shortcircuits the 1k2 resistor. This makes the output voltage of charger to go down to about 1.25V. If according comparator sees a voltage of less than about 11.3V then Px,x goes low (adjust the 10k trimmer accordingly). 11.3V is the minimum battery voltage that is needed to make the 9V regulator to work properly (check the correct level by yourself).

The lower comparator can be used to check the voltage coming from mains transformer. You can set the threshold to about 13.2V, then you can detect whether enough voltage comes to the 9V regulator, or you can set the threshold to about 15.5V, then you can detect whether enough voltage comes to fully charge the Pb-battery (decide yourself and check correct levels).

The switch allows you to turn-off the circuit powered by the 9V regulator, but keeping the Pb-battery connected to the charger.

Take care, both regulators should get a heatsink!

I know that there are much more elegant methodes to do the same, using chips from maxim and others. But why not trying it by the use of standard components?

Kai