How to bring a transformer's secondary impedance to the primary side with k < 1?
You need to calculate the magnetization inductance because otherwise, your question is flawed (as it imposes a belief that transformer impedance is governed purely by the load on the secondary and the turns ratio).
When you factor in the magnetization impedance you get a correct answer for k = 1 by making the reflected load impedance in parallel with this inductance.
When k is less than 1, you have leakage inductance (basically 1-k multiplied by the magnetization inductance) and, it is in series with the transformer primary and acts only on its own hence, it represents an impedance you have to add to the reflected impedance from the secondary. And, of course you have a reduced magnetization inductance due to k being less than one so, you also still have the same old parallel path to the reflected impedance and those two impedances are in series with the leakage inductance.