= Solution
There is an initial-value qualification in the printed conclusion. As written, nothing requires $A_0=0$. For example, $M\equiv0$ and $A\equiv1$ satisfy the hypotheses, because $M^2-A\equiv-1$ is a <local martingale>, but $M_t$ is not $N(0,1)$. \b[The correct variance is $A_t-A_0$; the printed formula holds under the usual normalization $A_0=0$.]
Indeed, $M^2-[M]$ is a <continuous local martingale>. Subtracting it from $M^2-A$ shows that $[M]-A$ is a continuous <local martingale> with <finite variation>, hence is its initial constant $-A_0$. Thus
$$
C_t:=A_t-A_0=[M]_t.
$$
In particular, $C$ is deterministic, continuous, nonnegative and increasing, regardless of the apparent allowance of a general <finite-variation process> in the premise.
For a real parameter $\theta$, the <Itô formula> gives the complex <local martingale>
$$
Z_t=\exp\left(i\theta M_t+\frac{\theta^2}{2}C_t\right),
\qquad dZ_t=i\theta Z_t\,dM_t.
$$
On every deterministic interval $[0,T]$, its modulus is at most $e^{\theta^2C_T/2}$, so its real and imaginary parts are bounded <local martingales>, hence true <martingales>. Taking <expectations> and using $Z_0=1$ yields
$$
\mathbb E e^{i\theta M_t}=e^{-\theta^2C_t/2}.
$$
By the <uniqueness theorem for characteristic functions>,
$$
\boxed{M_t\sim N(0,A_t-A_0).}
$$
A zero variance means the point mass at zero. This proves the requested normalized version by a direct <characteristic function> calculation.
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