Equalizers by pullbacks and products

Hint

Let's call $h: E \rightarrow B$. Now considering $B \times B$ from the definition know the existence of two maps $\pi_1, \pi_2 : B \times B \rightarrow B$ such that:

$\pi_1 \circ <f,g> = f$

$\pi_2 \circ <f,g> = g$

Knowing this we can see:

$f \circ e = \pi_1 \circ <f,g> \circ e = \pi_1 \circ \Delta \circ h = h= \pi_2 \circ \Delta \circ h = \pi_2 \circ <f,g> \circ e = g \circ e$