If $\mathscr A$ has all products and equalizers, then it has all limits This question is about part (a) of this proposition:

Here is a plan of the proof.

Here's what the picture looks like, from what I understand:

But I don't understand what the condition $s\circ p=t\circ p$ means. I guess this means componentwise equality: $s_u\circ p_u=t_u\circ p_u $ for all $u$, but what is $p_u$? $L$ is not a product, and so $p_u$ cannot be a projection.
I'm just trying to prove first that $Du\circ p_I=p_J$ for all $u: I\to J$, and it seems I need to use $s\circ p=t\circ p$ to that end, which is unclear how to do.

$s \circ p = t \circ p$ are both arrows into a product, the equality here is equivalent to them being equal after each projection, i.e., $\pi_u \circ s \circ p = \pi_u \circ t \circ p$ for all $u : J \to K$ in $\mathbf I$.
Now, $\pi_u \circ s$ is the $u$-component of $s$, hence is equal to $D(u) \circ \text{pr}_J$. Similarly, $\pi_u \circ t = \text{pr}_K$. Combining this with the previous equality, we find $D(u) \circ \text{pr}_J \circ p = \text{pr}_K \circ p$. Now, by definition $p_I = \text{pr}_I \circ p$ for any $I \in \mathbf I$ (i.e. $p_I$ is the $I$-component of $p$), we find $D(u) \circ p_J = p_K$.
Hence $(L, (p_I : L \to D(I))_{I \in \mathbf I})$ is a cone. Hopefully this clears up the confusion. 
