The categorical definitions of the basic objects (products, coproducts, equalizers, coequalizers, limits, colimits, etc) arise from abstracting a particular situation that shows up in many instances.
The notion of a "product" shows up repeatedly among Sets (cartesian product), Groups (cartesian product again), Rings (cartesian product yet again), Modules (and again), Topological spaces (and again).
Their "object" properties vary from category to category; in the class of abelian groups, the direct product contains the direct sum, which is an important objects in its own right (it's also important in Groups, but less so). Among Unital Rings, however, the direct sum is not an interesting object when it is different from the product (because it is not a unital ring). In topological spaces, there is a "natural" way of defining the topology on a product (the "box" topology), but it turns out to be lacking in some uneasy sense when there are too many factors in the product... and there is an alternative possibility (the "product topology") that seems to be behave better...
The general, guiding principle in Category Theory, though, is that what is important is not what an object "is", but rather what it "does": how it behaves relative to other objects in terms of arrows/morphisms. So we want to abstract the properties that all those instances share in $\mathbf{Sets}$, $\mathbf{Groups}$, $\mathbf{AbGroups}$, $\mathbf{Rings}$, $\mathbf{UnitalRings}$, $\mathbf{Semigroups}$, $\mathbf{TopSpaces}$, etc.
(In the latter case, it even suggests which of the two possible topologies "should" be used).
So... what do all these specific instances all have in common? From the cartesian product we can map to each factor: if we view the cartesian product $\times X_i$ as the set of functions $f\colon I\to \cup X_i$ with $f(i)\in X_i$ for each $i\in I$, then we have a natural map from the product to each $X_i$ by "evaluation": $\pi_i\colon f\longmapsto f(i)$. And $f$ is completely determined by these images; so that if you "select" $x_i\in X_i$ for each $i$, then this gives you a unique $\mathbf{x}\in \times X_i$.
Ah, but again: we don't want to think in terms of elements (what the objects "are"), but rather in terms of maps. For $\mathbf{Sets}$, selecting objects is the same as mapping from single element sets... but again we are worrying about the what the objects "are" (single element sets)...
Okay: since we care about maps, note this: not only can we go from $\times X_i$ to each $X_i$. These maps are such that if you have any set $Y$, and any maps $g_i\colon Y\to X_i$, then each $y\in Y$ determines an element of $\times X_i$, namely, $y\colon i\to g_i(y)$. Put another way:
If we let $\pi_i\colon\times X_i\to X_i$ be the "evaluate at $i$" map, then for any set $Y$ and any maps $g_i\colon Y\to X_i$, there is a way to map $g\colon Y \to \times X_i$ in such a way that $\pi_i\circ g(y) = g_i(y)$ for every $i$.
What's more, $g$ is completely forced by this condition. This is a good "categorical" property, because it is given entirely in terms of morphisms.
Does this property "go through" the other instances? The group with the coordinatewise product? The semigroup/ring/module with coordinatewise product? Yes! Good. What about topological spaces? Yes with respect to the product topology... no with respect to the Box topology... so we want to use the "product topology" for the "product".
This definition matches the many instances, and has nice "categorical" properties. It completely characterizes the product up to unique isomorphism. So it seems like a good definition. Turns out to be the right way to generalize the properties of the cartesian product/direct product to other categories.
One of the main properties of abelian categories is that every monomorphism is a kernel, and more specifically, the kernel of its cokernel; and dually, every epimorphism is the cokernel of its kernel.
In particular, $f$ factors through a monomorphism $i$ if and only if $\operatorname{Coker}(i)\circ f=0$; by the universal property of the cokernel $\operatorname{Coker}(f)$, this is also equivalent to the condition that there exists a (unique) morphism $h$ such that $h\circ \operatorname{Coker}(f) =\operatorname{Coker}(i)$. But since $\operatorname{Coker}(f)$ is also the cokernel of its kernel, the existence of $h$ is also equivalent to $\operatorname{Coker}(i)\circ \operatorname{Ker}(\operatorname{Coker}(f))=0$, and by the universal property of the kernel $i$ of $\operatorname{Coker}(i)$ this is equivalent to the existence of a morphism $g$ such that $\operatorname{Ker}(\operatorname{Coker}(f))=i\circ g$.
Best Answer
In your proof, you already have a contradiction as soon as you write $p_2 : A \to B$, since there is no morphism $A \to B$ in $\mathcal{C}$; you don't need the 'such that for any object W ...' part at all.
Similar contradictions can be obtained in the cases $P=B$ and $P=C$, since there is no morphism $B \to A$ and no morphism $C \to A$ in $\mathcal{C}$.