Martingale representation of European option. Let stock price $S$ satisfy
$$S(t)=S(0)e^{(\int_0^t\sigma(s)dB_s-\frac{1}{2}\int_0^t\sigma(s)^2ds)}$$
I want to calculate the Martingale representation $V(t)=E(F|F_t)$ of European option with strike price $M$ and maturity $T$ which is given by
$$F=(S(T)-M)^+$$
The problem should be able to be solved by the use of Ito-Ocone formula.

The Ito-Ocone formula is given by
$$E(F|F_t)-E(F)=\int_0^tE(D_sF|F_s)dB_s$$
where $F$ is of the form
$$F=f(\int_0^T h(s)dB_s)$$
and $f$ is a $C^2$, $h$ a $L^2$ function. Further $D_s$ is the Malliavin derivative.
$$D_sF=\nabla_x f(\int_0^T h(s)dB_s)h(s)$$

Question: How to compute $E(F|F_t)$, and find continuous smooth test
function such that $g(S_t,t,T)=E(F|F_t)$, and find partial differential equation satisfied by $g(x,t,T)$?
An attempt:
The functions can be chosen as:
$$f(x)=(e^{x-\frac{1}{2}\int_0^T\sigma(s)ds}-M)^+,h(t)=\sigma(t)$$
Then I get
$$E(D_sF|F_s)=\sigma(s)E(S(T)I(S(T)>M))|F_s)$$
And by the Ito-Clarck formula
$$E(F|F_t)=\int_0^t \sigma(s)E(S(T)I(S(T)>M))|F_s)dB_s+V(0)$$
I get stuck here, how to get the formula into a form such that I can find a smooth test function (presumably by using Ito formula)?
 A: Note that your Stock Market satisfies the following SDE
$$ dS(t) = \sigma(t)S(t)dB(t).$$
Hence it is a martingale with respect to $\mathcal{F}_t$ (the natural filtration of your Brownian motion $B$). 
The indicator function of an event $A$ will be denoted by $\mathbf{1}_A$.
We obtain:
\begin{align*}
\mathbb{E}[(S(T) - M)^+| \mathcal{F}_t]&= \mathbb{E}[S(T)\mathbf{1}_{S(T) \geq M}|\mathcal{F}_t] - M \mathbb{E}[\mathbf{1}_{S(T) \geq M} | \mathcal{F}_t] \\=&
\mathbb{E}[S(T)\mathbf{1}_{S(T) \geq M}|\mathcal{F}_t] - M \mathbb{P}(S(T) \geq M |  \mathcal{F}_t)
\end{align*}
The second term is easier to find so I leave it for later. For the first term, that is, $\mathbb{E}[S(T)\mathbf{1}_{S(T) \geq M}|\mathcal{F}_t]$ first use the following Girsanov's theorem 

Girsanov's Theorem
Let $T>0$. Suppose that for all $t \in [0, T]$, $W(t)$ is given by
$$dW(t) = g(t)dt + dB(t), \ W(0)=0$$
where $g\colon [0, T] \rightarrow \mathbb{R}$ is a continuous (deterministic) function. 
Then $$M_t^g = \exp \left\{ - \int_0^t g(s)dB(s) - \frac{1}{2} \int_0^t g^2(s) ds\right\}$$
is a martingale with respect to $\mathbb{P}$ and $\mathcal{F}_t$ and $W(t)$ is a standard Brownian motion with respect to the probability measure $\mathbb{Q}$, equivalent to $\mathbb{P}$, defined on $\mathcal{F}_T$ by
\begin{equation}\label{meas} d\mathbb{Q} = M_T^g(\omega)d\mathbb{P}(\omega).\end{equation}

Take $g(t)= - \sigma(t)$ and define a new Brownian motion $W$ by
$$ dW(t) = -\sigma(t)dt + dB(t)$$
with respect to measure $\mathbb{Q}$ defined as in the theorem with $M_t^g :=\frac{S(t)}{S(0)}$. 
Also note that with respect to $W$, the stock $S(t)$ has the following form:
$$S(t) = S(0)\exp\left\{ \frac{1}{2}\int_0^t \sigma^2(s)ds + \int_0^t \sigma(s)dW(s)\right\}$$
Hence, 
\begin{align*}
\mathbb{E}[S(T)\mathbf{1}_{S(T) \geq M}|\mathcal{F}_t]=&
S(0) \mathbb{E}[M_T^g \mathbf{1}_{S(T) \geq M} |\mathcal{F}_t] \\
=& S(0) \mathbb{E}_{\mathbb{Q}}[\mathbf{1}_{S(T) \geq M} |\mathcal{F}_t] \\
=& S(0)\mathbb{Q}(S(T) \geq M | \mathcal{F}_t).
\end{align*}
Now since $(S(t))$ is a Markov process for $\mathbb{P}$ and $\mathbb{Q}$, we have
\begin{align*}
\mathbb{P}(S(T) \geq M | \mathcal{F}_t)=&\mathbb{P}(S(T) \geq M |S(t))\\
\mathbb{Q}(S(T) \geq M | \mathcal{F}_t)=&\mathbb{Q}(S(T) \geq M | S(t))
\end{align*}
I will only find $\mathbb{P}(S(T) \geq M |S(t))$ and leave $\mathbb{Q}(S(T) \geq M |S(t))$ as an exercise. The only difference in finding them is that one has to use the closed form of $S(T)$ in the correct Brownian motion; namely, for finding the $\mathbb{P}$-term use $S(t)$ in $B$-BM-form, for $\mathbb{Q}$-term use $S(t)$ in $W$-BM-form. 
For $\mathbb{P}(S(T) \geq M | S(t))$, first note that
$$S(T) = S(t)S(t, T),$$
where $S(t, T) = \exp\left\{-\frac{1}{2} \int_t^T \sigma^2(s)ds + \int_t^T \sigma(s)dB(s) \right\}$, and $S(t, T)$ is independent from $S(t)$ (independent increment property of BM). Therefore, 
\begin{align*}
\mathbb{P}(S(T) \geq M | S(t)) =& \mathbb{P}\left(S(t,T) \geq \frac{M}{S(t)}|S(t)\right)\\
=&\mathbb{P}\left( \frac{ \int_{t}^T \sigma(s)dB(s)}{\sqrt{\int_t^T \sigma^2(s)ds}}\leq \frac{\ln\left(\frac{S(t)}{M} \right)- \frac{1}{2} \int_t^T \sigma^2(s)ds}{\sqrt{\int_t^T \sigma^2(s)ds}}\left| S(t)\right.\right)\\
=& \mathcal{N}\left( \frac{\ln\left(\frac{S(t)}{M} \right)- \frac{1}{2} \int_t^T \sigma^2(s)ds}{\sqrt{\int_t^T \sigma^2(s)ds}}\right), 
\end{align*}
where $\mathcal{N}$ is a CDF of standard normal distribution.
Above I used the fact that 
$$ \int_{t}^T \sigma(s)dB(s) \sim \mathcal{N}\left( 0, \int_t^T \sigma^2(s)ds\right).$$
