Answer
The corresponding particle is a $K^0$
Work Step by Step
The $\pi^-$ has strangeness $0$, charge $-1$, and baryon number $0$
The proton $p$ has strangeness $0$, charge $+1$, and baryon number $+1$
The left side of the process has net strangeness $0$, net charge $0$, and net baryon number $+1$
The $\Xi^0$ has strangeness $-2$, charge $0$, and baryon number $+1$
The $K^0$ has strangeness $+1$, charge $0$, and baryon number $0$
The $\Xi^0$ and the $K^0$ have net strangeness $-1$, net charge $0$, and net baryon number $+1$
According to the conservation laws, the $x$ must have strangeness $+1$, charge $0$, and baryon number $0$
The corresponding particle is a $K^0$
