Read the two sides of the junction from the fabrication step. The question describes an n-channel MOSFET made on a p-type substrate: the source and drain are formed by diffusing (or implanting) heavily doped n⁺ regions into that substrate.

One side of the junction is therefore the diffused region, doped n⁺ (donor concentration typically 10¹⁹–10²⁰ cm^−3); the other side is the substrate, which is moderately doped p (about 10¹⁵–10¹⁶ cm^−3). Naming the junction as (diffused region)/(substrate) gives

\(n^{+}/p\)

Why the notation matters. The superscript "+" denotes heavy doping, and the two sides are deliberately unequal. This makes it a one-sided junction, with two useful consequences:

1. The depletion region extends almost entirely into the lightly doped p-side, since charge neutrality requires \(N_A x_p = N_D x_n\) and \(N_D \gg N_A\).

2. The n⁺ region gives a low series resistance and forms a good ohmic contact to the metal, which is exactly why source and drain are doped so heavily.

Why the other options are wrong. n⁺/p⁺ would be a junction between two heavily doped regions — a tunnel-diode-like structure that breaks down at very low voltage, useless as a source/drain junction. n/p⁺ reverses the doping levels, and n/p describes two lightly doped sides, which is not what a diffused source/drain is.

Note on the device. These source/drain-to-body junctions are held reverse biased in normal operation so that they stay isolated from the substrate; their capacitance and leakage are what limit switching speed and standby current.

Hence, the junction formed is of the n⁺/p type.