Question regarding No Arbitrage price of a call option

I have a question regarding how to solve the NA price for a slightly modified call option.

Say that I have a money account $$B(T)=e^{r(T-t)}$$ and a stock dynamic $$\frac{dS(t)}{S(t)}=(r-\delta)dt+\sigma dW(t)$$ where r is the riskless return, $$\delta$$ the continous dividend yield and $$W$$ a brownian motion. By Itô's lemma we can easily derive $$S(T)=S(t)e^{r-\delta-\frac{\sigma^2}{2}+\sigma(W(T)-W(t))}$$.

I want now to compute the NA price for the T-claim $$X=max(B(T),S(T))$$. My solution so far is as follows below:

$$\Pi(t;X)=\frac{1}{B(T)}E^Q_t[max(B(T),S(T))]\\ =\frac{1}{B(T)}E^Q_t[B(T)+max(0,S(T)-B(T))]\\ =1+\frac{1}{B(T)}E^Q_t[max(0,S(T)-B(T))]$$

Last expression is a call option with strike $$B(T)$$. Can I then continoue to apply the Black-Scholes formula and write the NA price as following below?

$$\Pi(t;X)=1+N(d_1)\frac{S(T)}{B(T)}-N(d_2)$$

Or am I missing something? $$d_1$$ and $$d_2$$ is defined in the Black Scholes formula as,

https://en.wikipedia.org/wiki/Black%E2%80%93Scholes_model

Yes you can use the Black-Scholes Model with $$K=B(T)$$ because $$B(T)$$ is deterministic (a constant like $$K$$, since $$T$$ is constant).
However your current solution is incorrect as the Black-Scholes call price is already discounted ($$C:=e^{-rT}E^Q[(S_T-K)^+])$$ and based on the current stock price $$S_t$$ (not $$S_T$$). Further note that $$e^{-r(T-t)}B(T)=1$$ and $$1-N(x)=N(-x)$$:
\begin{align*}\Pi(t;X)&=1+N(d_1)S_te^{-\delta(T-t)}-N(d_2)e^{-r(T-t)}B(T)\\ &=N(d_1)S_te^{-\delta(T-t)}+N(-d_2) \end{align*}
Your solution is correct. Rewriting your modified payoff in terms of the payoff of a call option is a common technique. Note however that you have a little typo: you need $$S(t)e^{-\delta(T-t)}$$ instead of $$S(T)$$ in the last line, i.e. \begin{align*} \Pi(t,X) &= 1+S_te^{-\delta(T-t)}N(d_1)-N(d_2) \\ &= S_te^{-\delta(T-t)}N(d_1) +N(-d_2), \end{align*} since $$1-N(d_2)=N(-d_2)$$. In fact, we have $$\mathbb{E}^{\mathbb{Q}}_t[S(T)]=S(t)e^{-\delta(T-t)}$$. Furthermore, you can simplify $$d_1$$ and write \begin{align*} d_1 &= \frac{\ln\left(\frac{S(t)}{B(T)}\right)+\left(r-\delta+\frac{1}{2}\sigma^2\right)(T-t)}{\sigma\sqrt{T-t}} \\ &= \frac{\ln\left(S(t)\right)-r(T-t)+\left(r-\delta+\frac{1}{2}\sigma^2\right)(T-t)}{\sigma\sqrt{T-t}} \\ &= \frac{\ln\left(S(t)\right)-\left(\delta-\frac{1}{2}\sigma^2\right)(T-t)}{\sigma\sqrt{T-t}} \end{align*}