A splitting of an idempotent is simultaneously an equalizer and a coequalizer and is unique up to unique isomorphism

Since $e^2=e$ and $ip=e$, one has $ei=ipi=i=1_{A}i$. If $h:X\to A$ also satisfies $eh=h$, then $i(ph)=iph=eh=h$. If $iu=h$ too, then $u=piu=ph$. So $i$ is the equalizer of $e$ and $1_A$.

Dually, $pe=pip=p$, so $p$ coequalizes $e$ and $1_A$. If $h:A\to X$ satisfies $he=h$, then $(hi)p=hip=he=h$, and uniqueness follows because any $u:B\to X$ with up=h must satisfy $u=upi=hi$. Thus $p$ is the coequalizer.

Let $A\stackrel{p^{\mathrm{\prime }}}{\to }{B}^{\mathrm{\prime }}\stackrel{i^{\mathrm{\prime }}}{\to }A$ be another splitting of $e$. Put $u:=p^{\mathrm{\prime }}i:B\to B^{\mathrm{\prime }}$ and $v:=p{i}^{\mathrm{\prime }}:B^{\mathrm{\prime }}\to B$. Then $i^{\mathrm{\prime }}u=i^{\mathrm{\prime }}{p}^{\mathrm{\prime }}i=ei=i$ and $vp=p{i}^{\mathrm{\prime }}{p}^{\mathrm{\prime }}=pe=p$, so these are the unique morphisms commuting with the legs. Moreover $vu=p{i}^{\mathrm{\prime }}{p}^{\mathrm{\prime }}i=pei=pi=1_B$ and $uv=p^{\mathrm{\prime }}ip{i}^{\mathrm{\prime }}=p^{\mathrm{\prime }}e{i}^{\mathrm{\prime }}=p^{\mathrm{\prime }}{i}^{\mathrm{\prime }}=1_{B^{\mathrm{\prime }}}$. Hence $u$ is the unique isomorphism between the two splittings by [L3].