I use only the fact that
$S(t)\cdot B^{-1}(t)$ is a $\mathbb{Q}-$ martingale and we are considering the European options.
Indeed,
We have that : $(x-K)_{+}-(K-x)_{+}=x-K$ $\forall x,K \in \mathbb{R}$ ($\diamond$)
But we know also that the price of a European Call/Put is given by :
$C(t) = \mathbb{E}^{\mathbb{Q}}\left[\left(S_T-K\right)_{+}\cdot \frac{B(t)}{B(T)}\, \mid \, \mathcal{F}_{t}\right]$ , $P(t) = \mathbb{E}^{\mathbb{Q}}\left[\left(K-S_T\right)_{+}\cdot \frac{B(t)}{B(T)}\, \mid \, \mathcal{F}_{t}\right]$
Using the ($\diamond$), we have
$C(t)-P(t) = \mathbb{E}^{\mathbb{Q}}\left[\left(S_T-K\right)\cdot \frac{B(t)}{B(T)}\, \mid \, \mathcal{F}_{t}\right]=B(t)\cdot \mathbb{E}^{\mathbb{Q}}\left[ \frac{S_T}{B(T)}\, \mid\,\mathcal{F}_{t}\right]-\frac{B(t)}{B(T)}\cdot K$
Using the assumption, we have $B(t)\cdot \mathbb{E}^{\mathbb{Q}}\left[ \frac{S_T}{B(T)}\, \mid\,\mathcal{F}_{t}\right]=B(t)\times \frac{S_{T}}{B(t)} = S_{t}$
We also know that $\frac{B(t)}{B(T)}=B(t,T)$ (I think if not, there's arbitrage)
At the end, we get the result :)