As I said in a comment, you could use a mixer to do the work in the analog domain. You could use a chip such as a AD831 which work from DC to several MHz. It also contains some OP amps in the same package that you don't need to use.

This works as follow:

Use a waveform generator to get a sine at a defined frequency 100kHz for instance. Enter it to the fist input of the mixer and also to the sensor input.

Connect the sensor output to the second input of the mixer.

Add a basic low pass filter to the mixer output with a cutoff frequency a lot lower than the sine frequency (let say here at 1kHz).

How it works ? the mixer is a multiplier. its ideal response is equal to :

$$ V_{out} = A_1cos(\omega_1t)A_2cos(\omega_2t) $$

But here the input frequencies are the same and omega 1 = omega 2 only a phase shift phi exists : $$ V_{out} = A_1cos(\omega t)A_2cos(\omega t + \phi) $$

which can be rewritten : $$ V_{out} = \frac{A_1 A_2}{2} [cos(2\omega t+\phi) + cos( \phi) ]$$ Here we have one therm that changes over time (the one with the 2wt+phi) and one that is constant (the phi alone). Thus,the lowpass filter at the output of the mixer will remove the oscillating term and the resulting voltage only depends of three factors  :

1. The input amplitude A1 which is fixer here.
2. The amplitude at the sensor output. You have to check if it changes significantly. Probably not that much.
3. The phase (phi) that you want to measure.

The output voltage after the mixer and the filter is a "DC" voltage that depends of the phase you want to measure. You don't need to have a fast ADC here. You only measure a quasi DC signal.

As I said in a comment, you could use a mixer to do the work in the analog domain. You could use a chip such as a AD831 which work from DC to several MHz. It also contains some OP amps in the same package that you don't need to use.

This works as follow:

Use a waveform generator to get a sine at a defined frequency of 100kHz for instance. Enter it to the fist input of the mixer and also to the sensor input.

Connect the sensor output to the second input of the mixer.

Add a basic low pass filter to the mixer output with a cutoff frequency a lot lower than the sine frequency (let say here at 1kHz).

How it works ? the mixer is a multiplier. Its ideal response is equal to :

$$ V_{out} = A_1cos(\omega_1t)A_2cos(\omega_2t) $$

But here the input frequencies are the same and omega 1 = omega 2 only a phase shift phi exists : $$ V_{out} = A_1cos(\omega t)A_2cos(\omega t + \phi) $$

which can be rewritten : $$ V_{out} = \frac{A_1 A_2}{2} [cos(2\omega t+\phi) + cos( \phi) ]$$ Here we have one therm that changes over time (the one with the 2wt+phi) and one that is constant (the phi alone). Thus, the lowpass filter at the output of the mixer will remove the oscillating term (because its mean is equal to zero) and the resulting voltage only depends of three factors:

1. The input amplitude (A1) which is fixer here.
2. The amplitude at the sensor output (A2). You have to check if it changes significantly along the scale of the sensor. Probably not that much.
3. The phase (phi) that you want to measure.

The output voltage after the mixer and the filter is a "DC" voltage that depends of the phase you want to measure. You don't need to have a fast ADC here. You only measure a quasi DC signal.

I don't have here any tool to draw a schematics. Sorry.