= Solution
Let $P=m_pc^2$ denote the proton rest energy and $W=220\,\mathrm{GeV}$ the desired particle's rest energy. For an antiproton of total <energy> $E_{\bar p}$ striking a stationary proton, the incoming squared <centre-of-mass energy> is
$$
s=(E_{\bar p}+P)^2-c^2p_{\bar p}^2=2P^2+2PE_{\bar p}.
$$
For production of the stated single particle, <four-momentum conservation> requires $s=W^2$. Thus
$$
\boxed{E_{\bar p}^{\rm fixed}=\frac{W^2-2P^2}{2P}\simeq25.7912\,\mathrm{TeV}.}
$$
The numerical value uses $P=0.938272\,\mathrm{GeV}$. It is a \b[total] incident <energy>; the corresponding <kinetic energy> is $E_{\bar p}^{\rm fixed}-P\simeq25.7902\,\mathrm{TeV}$. Much of the laboratory <energy> supplies motion of the outgoing system rather than its <invariant mass>.
For equal and opposite beam <momenta>, the laboratory is the <centre-of-mass frame>. Each beam has the same total <energy> and their sum is $W$, giving
$$
\boxed{E_{\bar p}^{\rm collider}=\frac W2=110\,\mathrm{GeV},\qquad
K_{\bar p}^{\rm collider}=110-P\simeq109.062\,\mathrm{GeV}.}
$$
These are the <fixed-target production threshold> and collider kinematic comparison for the specified final state. Extra final particles or restrictions from conserved quantum numbers would require a separately specified reaction.
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