The intuitive explanation is given in @Alex C's comment. You should stick to that if you understand it.
Yet, if you are more comfortable with a mathematical approach:
- Payoff of being long a forward contrat with maturity $T$: $(S_T - X)$. Interpretation: at time $T$, you pay a certain price $X$ in exchange for which you receive the underlying $S_T$
- Payoff of being long a call option struck at $X$ and with maturity $T$: $(S_T - X)^+$. Interpretation: at time $T$, you exercise if the underlying's value is greater than the strike price, in which case you earn the difference between the 2, otherwise you do nothing.
- Payoff of being long a put option struck at $X$ and with maturity $T$: $(X - S_T)^+$. Interpretation: at time $T$, you exercise if the underlying's value is smaller than the strike price, in which case you earn the (absolute) difference between the 2, otherwise you do nothing.
- Payoff of being long (+) call option struck at $X$ and short (-) a put struck at $X$ at maturity $T$, is the difference of the 2 previous ones:
\begin{align}
(S_T - X)^+ - (X - S_T)^+ &= (S_T - X)1\{S_T \geq X\} - (X - S_T)1\{S_T \leq X\} \\
&= (S_T - X)1\{S_T \geq X\} + (S_T - X)1\{S_T \leq X\} \\
&= (S_T - X)
\end{align}
As you can see, being long a forward with 'strike' $X$ and being long a call/short a put with strikes $X$ give the same payoff at $T$: $(S_T-X)$.
By absence of arbitrage opportunity, these two strategies should therefore have exactly the same value today, that is:
$$ C_E(S_0;T,X) - P_E(S_0;T,X) = V_X(0) $$
which is the famous put-call parity relationship, where $V_X(0)$ represents the value of being long a forward contract struck at $X$ and with maturity $T$ as seen of today, in other words
$$ V_X(0) = P(0,T)(F(0,T)-X)$$
with $P(0,T)$ representing the discount factor applying to cash-flows paid at $T$ and $F(0,T)$ the fair forward value.
By setting $X$ (strike price) equal to $F(0,T)$ (forward value $S_0e^{rT}$ in the absence of dividends), then you see that $V_X(0) = 0$ and put-call parity re-writes as
$$ C_E(S_0;T,X) - P_E(S_0;T,X) = V_X(0) = 0 $$
meaning
$$ C_E = P_E $$