= Solution
An \b[<abelian category>] is an <additive category> with a <kernel in a category> and a <cokernel in a category> for every morphism, in which every <monomorphism> is a kernel and every <epimorphism> is a cokernel. In particular, every <epimorphism> $p$ is the cokernel of its own kernel: if $p$ is initially a cokernel of $u$, then $u$ factors through $\ker p$, and the two cokernel universal properties agree. The dual statement holds for <monomorphisms>. Finite <biproducts> and kernels construct <finite limits>.
For <pullback stability of epimorphisms in an abelian category>, let $p:B\to C$ be an <epimorphism> and $k:C'\to C$ any morphism. The map
$$
d=[p,-k]:B\oplus C'\to C
$$
is epic, since its restriction to the $B$ summand is $p$. Its <kernel in a category> $i=(h,p'):B'\to B\oplus C'$ exhibits the <pullback in a category>: the equation $d(h,p')=0$ means $ph=kp'$ and has exactly that universal property.
Suppose $v:C'\to D$ satisfies $vp'=0$. The morphism $[0,v]:B\oplus C'\to D$ annihilates $i$, so, because $d$ is the <cokernel in a category> of $i$, it factors as $[0,v]=wd$ for some $w:C\to D$. Restricting to $B$ gives $wp=0$, and epicity gives $w=0$. Hence $v=0$. Applying this to the difference of any two morphisms agreeing after $p'$ proves that $p'$ is an <epimorphism>. Therefore
$$
\boxed{\text{Epimorphisms in an abelian category are stable under pullback}.}
$$
This proof uses normality from the definition, rather than assuming the stability to be proved.
For the <pullback of a short exact sequence in an abelian category>, use the displayed square's notation $g,g',h,k$. Exactness makes $f:A\to B$ the <kernel in a category> of $g$. Since $gf=0=k0$, the <pullback in a category> gives a unique
$$
f':A\to B',\qquad hf'=f,\qquad g'f'=0.
$$
If $u:X\to B'$ has $g'u=0$, then $g(hu)=kg'u=0$, so there is a unique $v:X\to A$ with $fv=hu$. Both projections of the pullback give $f'v=u$: their composites with $h$ are equal and their composites with $g'$ are zero. The uniqueness of $v$ follows from monicity of $f$. Thus $f'$ is the <kernel in a category> of $g'$. By the preceding stability result, $g'$ is an <epimorphism>. We obtain the required <exact sequence in an abelian category>:
$$
\boxed{0\longrightarrow A\xrightarrow{f'}B'\xrightarrow{g'}C'\longrightarrow0.}
$$
Finally form the <pullback in a category> $E=P\times_CP'$ of the two projective presentations. Pulling back each of their <short exact sequences in an abelian category> gives
$$
0\longrightarrow K\longrightarrow E\longrightarrow P'\longrightarrow0,
\qquad
0\longrightarrow K'\longrightarrow E\longrightarrow P\longrightarrow0.
$$
Because each of $P'$ and $P$ is a <projective object in a category>, the respective final <epimorphisms> have sections: lift their identity morphisms through those epimorphisms. The usual <additive category> splitting argument identifies each middle object with the <biproduct> of its kernel and quotient. Explicitly, a kernel inclusion $j$ and section $s$ give the isomorphism $[j,s]$ from that biproduct to the middle object; $1-sq$ factors through $j$ and provides its inverse's kernel component. Hence
$$
E\cong K\oplus P',\qquad E\cong K'\oplus P,
$$
and we conclude the \b[<Schanuel lemma in an abelian category>]:
$$
\boxed{K\oplus P'\cong K'\oplus P.}
$$
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