Extremely high impedance peaks like that can often be well addressed by a high value resistor in parallel with the inputs
I hope you know what you are talking about here. A “high-value” parallel resistor — say, 100 Ω — won’t effectively flatten an impedance peak that high. The resulting impedance would still be around 43 Ω, based on 1/(1/75 + 1/100). That peak wouldn’t come close to the nominal impedance until a much lower value, such as 10 Ω, is used — which yields 1/(1/75 + 1/10) = 8.8 Ω.
Also, there is a downside of using a single parallel resistor, i.e., it affects the entire impedance spectrum, not just the peak. Even the region already near nominal impedance would drop — for example, 1/(1/8 + 1/10) = 4.4 Ω — effectively turning an easy-to-drive speaker into a more demanding one.
If the goal is to address only the impedance rise around 2 kHz (as MoFi did with the SP-10 ME), a more refined approach is required. I’d be happy to explain how that can be done if you could not figure that out, ’Engineer.’