Connecting an active low Enable pin IC with a MCU
1.1 MΩ is a high resistance, making the internal pull-up resistor very weak. By connecting an external pull-down resistor of the same value, you have effectively created a voltage divider such that the pin will, absent noise or unwanted signals, be indeterminate.

Your schematic is correct, but the value of pull-down resistor should be re-evaluated. In order to assert a definitely low condition on the EN_N pin, consider a stronger 10 KΩ value.

(If battery life is a concern, higher resistance values may be warranted, but must be low enough compared to the internal pull-up to produce a sufficiently "low" voltage.)

Consider:

$$V_{out} = \frac{V_{source} \times R_2}{(R_1 + R_2)}$$



Simple voltage divider (CircuitLab is not cooperating today)

If R1 is the internal pull-up and R2 is the external pull-down, then with both values equal you have 50% of V_source. For example if V_source is 3.3V, your EN_N pin would "see" 1.65V, which would not be considered "low."

If R2 is 500KΩ, V_out becomes approximately 1V, which might be considered low depending on the IC.

10K to 100K would be my recommended range but depends on the IC, noise immunity, tolerance for constant current consumption, etc.