As Luc points in the comments, $\bar u, u$ are row and column vectors, so $\bar u \gamma^\mu u$ is a scalar. Thus the expression may be arranged this way: $$\frac{e^4}{(p_3-p_1)^4}\underbrace{[\bar u(p_3)\gamma^\mu u(p_1)]}_1\underbrace{[\bar u(p_4)\gamma_\mu u(p_2)]}_2\underbrace{[\bar u(p_2)\gamma_\nu u(p_4)]}_3\underbrace{[\bar u(p_1)\gamma^\nu u(p_3)]}_4.$$ Changing the order from $1234$ to $1423$ gives Zee's result.

As Luc points in the comments, $\bar u, u$ are row and column vectors, so $\bar u \gamma^\mu u$ is a scalar. Thus the expression may be arranged this way: $$\frac{e^4}{(p_3-p_1)^4}\underbrace{[\bar u(p_3)\gamma^\mu u(p_1)]}_1\underbrace{[\bar u(p_4)\gamma_\mu u(p_2)]}_2\underbrace{[\bar u(p_2)\gamma_\nu u(p_4)]}_3\underbrace{[\bar u(p_1)\gamma^\nu u(p_3)]}_4.$$ Changing the order from $1234$ to $1423$ gives Zee's result.

Edit: As @JamalS points out, $\bar u, u $ carry spinor indices and $\gamma^\mu$ a Lorentz index, so $\bar u\gamma^\mu u$ is a scalar with respect to spinor indices only.