Looks good to me.
Here’s the model I’ve tried with identical Parallel RLC branches having values: R = 1e9 Ohm, L = (1e-1)/(2*pi) H, C = 1e-12 F.
Here are the results I’ve got for the parallel case and singular case:
Both these results match theoretical expectations.
It reads:
If you plan to connect the Impedance Measurement block in series with an inductance, a current source, or any nonlinear element, you must add a large resistor across the terminals of the block, because the Impedance Measurement block is simulated as a current source block.
I have attached the model I used to verify this with this Answer. Have a look.