Step 1: Track quark content instead of naming the particles.
Write each particle in the reaction using its quark content: $p = uud$, $n = udd$, $\pi^- = \bar{u}d$, $\pi^0 = \frac{1}{\sqrt{2}}(u\bar{u} - d\bar{d})$, $\Lambda^0 = uds$, $\bar{p} = \bar{u}\bar{u}\bar{d}$. Checking conservation this way makes the bookkeeping concrete rather than abstract.
Step 2: Redo option (A) at the quark level.
$\pi^- + p = (\bar{u}d) + (uud)$ gives quark content $u, u, d, d, \bar u$ (one $u\bar u$ pair can annihilate). $\pi^0 + n$ gives $u\bar u$ (or $d\bar d$) plus $udd$. Both sides carry the same net quark numbers, zero net strangeness, and zero net charge. Nothing blocks this rearrangement through the strong force, so it goes through.
Step 3: Redo option (B) using photon counting instead of quarks.
$\pi^0$ decay to photons is an electromagnetic process, and electromagnetism conserves C-parity exactly. The measured $\pi^0$ has $C=+1$; a state of $n$ real photons carries $C=(-1)^n$. Two photons give $C=+1$ (the real decay mode), but three photons give $C=-1$, a mismatch. A mismatch in a conserved quantum number blocks the process outright, regardless of how much phase space is available.
Step 4: Redo option (C) by counting net quarks minus antiquarks.
For $p\bar p$, net (quark minus antiquark) count is $3 - 3 = 0$. For $\Lambda^0 + \Lambda^0 = (uds) + (uds)$, net count is $6 - 0 = 6$. These do not match, so this transition is impossible; nature instead allows $p\bar p \to \Lambda^0\bar\Lambda^0$, where the second $\Lambda^0$ is replaced by its antiparticle so the antiquarks balance out.
Step 5: Redo option (D) with a simple energy-momentum count.
A lone photon final state has only its energy and the fixed relation $E = pc$ to work with, while the two-body initial state at rest supplies a fixed total energy and zero total momentum; one photon cannot carry zero momentum and non-zero energy at the same time, so this final state has no consistent solution and is forbidden.
Final Answer:
Quark-level and quantum-number bookkeeping both single out the same reaction as consistent.\[ \boxed{\pi^- + p \to \pi^0 + n} \]