Step 1: Photon energy
Each photon carries energy $h\nu$, fixed by the source frequency, not by its distance.
Step 2: What intensity changes
Lower intensity means fewer photons per second, so the photocurrent falls.
The maximum kinetic energy of an emitted electron $K_{max}=h\nu-\phi$ stays the same.
Step 3: Result
$V_0=\dfrac{K_{max}}{e}$ is unchanged, so option (C) is correct.
Final Answer:
The stopping potential stays the same.
\[ \boxed{\text{(C) }\text{will remain constant}} \]