Suppose $C$ is a category and $(X_i)_{i\in I}$ is a collection of objects.
Suppose $\Pi_{i\in I} X_i$ exists in $C$. If $a$ is another object in $C$. Is it possible to define a morphism $\Pi_{i\in I} X_i \rightarrow a$ from a collection of maps $f_i:X_i\rightarrow a$? I.e. is the arrow $\phi:\Pi_{i\in I} X_i\rightarrow a$ defined by $\pi_i\circ \phi = f_i$ a valid arrow?
No, this is not possible in general (up to a few reinterpretations I had to do). We can in fact give a simple counterexample with sets. Consider the product $X \times Y$ (where $X$ and $Y$ are non-empty), and suppose that $A = \{0, 1\}$ (I replaced your notation for $a$ with the capital letter, for a bit more consistency in notation). Then we can define $f_X: X \to A$ to send everything to $0$ and $f_Y: Y \to A$ to send everything to $1$.
In your question you want to compose an arrow $\phi: X \times Y \to A$ with the projections $\pi_X$ and $\pi_Y$. That is not possible (just check the domains and codomains). So I will interpret it in the only way that is possible: we will want that $\phi = f_X \pi_X$ (and similarly for $Y$).
Now suppose there is a function $\phi: X \times Y \to A$, such that $\phi = f_X \pi_X$ and $\phi = f_Y \pi_Y$. Then for $(x, y) \in X \times Y$ we have $$ 0 = f_X(x) = f \pi_X(x, y) = f \pi_Y(x, y) = f_Y(y) = 1, $$ which is a contradiction. So such $\phi$ cannot exist.