The notation $\mathbb{Z}[i]/(p)$ means $\mathbb{Z}[i]$ modulo the ideal generated by $p$, i.e., the multiples of $p$ (in the Gaussian integers).
The expression on the right is supposed to be $\mathbb{Z}_p[x]/(x^2+1)$. So first, $\mathbb{Z}_p$ is the integers modulo $p$, $\mathbb{Z}/p\mathbb{Z}$. Then you take $\mathbb{Z}_p[x]$, which are the polynomials with coefficients in $\mathbb{Z}_p$. Finally, you take the quotient modulo $x^2+1$, which means two polynomials with coefficients in $\mathbb{Z}_p$ are equivalent if and only if their difference is a multiple of $x^2+1$ (multiple in $\mathbb{Z}_p[x]$).
If you know the Isomorphism Theorem for rings, then the simplest way to do this is to construct a ring homomorphism either from $\mathbb{Z}[i]$ to $\mathbb{Z}_p[x]/(x^2+1)$ that is onto, and has kernel equal to the multiples of $p$; or a ring homomorphism from $\mathbb{Z}_p[x]/(x^2+1)$ to $\mathbb{Z}[i]/(p)$ that is onto and whose kernel are exactly the multiples of $x^2+1$. The latter is probably simpler.
If you don't know the Isomorphism Theorem for rings, you can try to define a homomorphism directly. You want to decide how to map an equivalence class of the form $a+bi + (p)$ ($a,b\in\mathbb{Z}$) to an equivalence class of the form $\overline{r}+\overline{s}x + (x^2+1)$, where $\overline{r}$ and $\overline{s}$ are integers modulo $p$ (you should first prove that every element in $\mathbb{Z}_p[x]$ is equivalent modulo $x^2+1$ to a polynomial of degree at most $1$; try using the division algorithm). There is an obvious choice: map $a+bi + (p)$ to $\overline{a}+\overline{b}x + (x^2+1)$ (the reason you would want to do this is that $x+(x^2+1)$ satisfies $t^2 + 1 = 0$ in $\mathbb{Z}_p[x]/(x^2+1)$; that is, it is a "square root of $-1$", just like $i$). You would need to show this is well-defined, one-to-one (on classes), onto, and a ring homomorphism.