There is a presentation by Michael D. Hirschhorn from University of Shanghai, you can find by searching "Simple proofs of Ramanujan's partition congruences."
It states: $$\sum_{n\geq0} p(n)q^n = \frac{1}{E}=\frac{E^4}{E^5}=\frac{EJ}{E(q^5)}=\frac{(E_0+E_1+E_2)(J_0+J_1)}{E(q^5)} =\frac{E_0J_0+E_0J_1+E_1J_0+E_1J_1+E_2J_0+E_2J_1}{E(q^5)},\tag{1}$$ where \begin{gather*} E=(1-q)(1-q^2)(1-q^3)\cdots=1-q-q^2+q^5+q^7-q^{12}-q^{15}+\cdots\tag{2}\\ J=E^3 \tag{3}\\ (1-q)^5\equiv 1-q^5 \pmod{5} \tag{4} \end{gather*} and $E_i, J_i$ consist of exponents with $I$ modulo $5$. It further says
...if we look for terms in which the exponent is congruent to $4$ modulo $5$, we find $$\sum_{n\geq0} p(5n+4)q^{5n+4} \equiv 0 \pmod{5} \tag{5}$$
My question is: How did they deduce (5) out of last row of (1)?
P.S.
I am only collecting Ramanujan's work for my high school so I do not really need to understand this, but everything else is clear to me except this.
Here's what the notation means. If $$A=A(q)=a_0+a_1q+a_2q^2+\cdots$$ then $A_j$ consists of the terms of $A$ where the exponents are congruent to $j$ modulo $5$. Therefore $$A_3=a_3q^3+a_8q^8+a_{13}q^{13}+\cdots$$ etc.
You are confusing equalities and congruences here. First of all $$J(q)=E(q)^3=\sum_{m=0}^\infty (-1)^m (2m+1)q^{m(m+1)/2}.$$ The exponents are triangular numbers, and these are congruent to $0$, $1$, $3$ modulo $5$, so that $$J=J_0+J_1+J_5.$$ However, if $5\mid m (m+1)/2$ then $5\mid (2m+1)$ so that the coefficients of $J_3$ are all multiples of $5$. So we can write $$J\equiv J_0+J_1\pmod 5$$ interpreted as the difference of the two sides has coefficient divisible by $5$.
Euler's formula for $E(q)$ implies that $E=E_0+E_1+E_2$. Therefore $$E^4=EJ\equiv(E_0+E_1+E_2)(J_0+J_1)\pmod 5$$ and the RHS of this is a power series with zero coefficients for the $q^m$ with $m\equiv4\pmod5$. Therefore the coefficient for such $q^m$ in $E_4$ are divisible by $5$, equivalently $(E^4)_4 \equiv0\pmod p$.
We have $P(q)=\sum p(n)q^n=1/E=E^4/E^5$. Then $E(q)^5\equiv E(q^5)\pmod 5$ (true for any power series), so $1/E^5\equiv (1/E^5)_0\pmod 5$. Then $$P\equiv((E^4)_0+(E^4)_1+(E^4)_2+(E^4)_3)(1/E^5)_0\pmod 5$$ giving $$P_4\equiv0\pmod 5.$$