This is a question due to the answer of Did in this post Independent increments of $X_t:=\int_0^t\phi(s) dW_s$. Precisely, we assume that the dynamics of a stock prices follows
$$dS_t=S_t(\phi_tdW_t)$$
where $\phi_t$ is a deterministic function, and assume bounded. Hence $S_t=\mathcal{E}(\phi\bullet W)=S_0e^{\int_0^t\phi_sdW_s-\frac{1}{2}\int_0^t\phi_s^2ds}$. For simplicity, $S_0=1$ and, regarding the question above, $X_t:=\int_0^t\phi_sdW_s$. Let $K>0$ be a constant, and we want to find the arbitrage free price of an European call option with strike $K$ and maturity $T$ at time $t$, i.e.
$$\pi_t:=E[(S_T-K)^+|\mathcal{F}_t]$$
As usual, with $A=\{S_T>K\}$ write this as
$$\pi_t=E[S_T\mathbf1_A|\mathcal{F}_t]-KE[\mathbf1_{A}|\mathcal{F}_t]$$
The second term can be easily calculated, using the above question, i.e.
$$KE[\mathbf1_{A}|\mathcal{F}_t]=KP[\frac{S_T}{S_t}>\frac{K}{S_t}|\mathcal{F}_t]$$
since $\frac{S_T}{S_t}$ is independent of $\mathcal{F}_t$ and $\frac{K}{S_t}$ is $\mathcal{F}_t$ measurable, we get
$$KP[\frac{S_T}{S_t}>\frac{K}{S_t}|\mathcal{F}_t]=KP[Y>\frac{K}{x}]|_{x=S_t}$$
where $Y$ is lognormal distributed with mean $-\frac{1}{2}\int_t^T\phi_s^2 ds$ and variance $\int_t^T\phi_s^2ds$. To calculate the first conditional expectation, $E[S_T\mathbf1_A|\mathcal{F}_t]$, I would define a new measure (on $\mathcal{F}_T$ through
$$\frac{dQ}{dP}:=S_T$$
since $S_T$ is a $P$ Martingale and $E[S_0]=1$ (if $S_0\not=1$, just use the measure $\frac{dQ}{dP}:=\frac{S_T}{S_0}$). Hence the density process is given by $Z_t:=E[\frac{dQ}{dP}|\mathcal{F}_t]=S_t$. Using Bayse, we get
$$S_tE[\frac{S_T}{S_t}\mathbf1_A|\mathcal{F}_t]=S_tE_Q[\mathbf1_A|\mathcal{F}_t]$$.
If we know that under $Q$, $\frac{S_T}{S_t}$ is still independent of $\mathcal{F}_t$, we could conclude in the same way as above. If this is not the case, how would you then calculate the first conditional expectation?
...simply note (as you already explained yourself) that $S_T=YS_t$ where the distribution of $Y$ is lognormal with known parameters, and $Y$ is independent of $\mathcal F_t$.
Since $A=[S_T\gt K]=[YS_t\gt K]$, this remark yields the identity $$\mathbb E[S_T\mathbf 1_A|\mathcal{F}_t]=\mathbb E[YS_t\mathbf 1_{YS_t\gt K}|\mathcal{F}_t]=G(S_t), $$ where the function $G$ is defined by $$ G(x)=\mathbb E(Yx\mathbf 1_{xY\gt K})=x\mathbb E(Y\mathbf 1_{Y\gt K/x}). $$