= Solution
Since $P(T,T)=1$, part (b) shows that the <density process> of the <T-forward measure> is
$$
L_t^T=\mathbb E^{\mathbb Q}\!\left[\frac{D_T}{P(0,T)}\,\middle|\,\mathcal F_t\right]
=\frac{D_tP(t,T)}{P(0,T)}
=\mathcal E\!\left(-\int_0^\cdot b_s^T\,dW_s\right)_t.
$$
Its terminal <expectation> is one and it is strictly positive, so the stated <Radon-Nikodym derivative> defines an <equivalent probability measure>. By the <Girsanov theorem>,
$$
W_t^T=W_t+\int_0^t b_s^T\,ds
$$
is a <Brownian motion> under $\mathbb Q_T$. Substitution of $dW_t=dW_t^T-b_t^Tdt$ cancels the entire <instantaneous forward rate> drift:
$$
df(t,T)=\sigma(t,T)\,dW_t^T.
$$
Because the integrand is deterministic and bounded, its <Itô integral> is square-integrable on $[0,T]$. With the usual fixed initial <instantaneous forward rate> curve, \b[the requested true martingale is]
$$
\boxed{f(t,T)=f(0,T)+\int_0^t\sigma(s,T)\,dW_s^T.}
$$
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