= Solution
The unit <zero-coupon bond> pays one at maturity. Its <risk-neutral valuation> therefore is
$$
\boxed{P(t,T)=\mathbb E_{\mathbb Q}\left[\exp\left(-\int_t^Tr_sds\right)\mid r_t=r\right].}
$$
No particular rate coefficients were specified, so this expectation is the general answer; evaluating it further requires a chosen <one-factor short-rate model>.
For the alternative derivation let $D_t=\exp(-\int_0^tr_sds)$ be the bank <discount factor>. For an integrable discounted payoff, pricing gives
$$
D_tV(t,r_t)=\mathbb E_{\mathbb Q}[D_TX\mid\mathcal F_t],
$$
which is a <martingale> by the <tower property of conditional expectation>. For a payoff determined by the <Markov> rate state, this conditional expectation is represented by the indicated pricing function. A path-dependent claim requires an appropriately enlarged state, as in part (c). Now $dD_t=-r_tD_tdt$ has finite variation. Applying <Itô formula> and the product rule under the <risk-neutral measure>, whose rate drift is $b-\sigma\theta$, yields
$$
d(D_tV)=D_t\left[V_t+(b-\sigma\theta)V_r+\frac12\sigma^2V_{rr}-rV\right]dt+D_t\sigma V_r\,d\widetilde W_t.
$$
Under the smoothness and integrability conditions used above, the <martingale> has zero finite-variation drift. The bracket therefore vanishes, giving again
$$
\boxed{V_t+(b-\sigma\theta)V_r+\frac12\sigma^2V_{rr}-rV=0.}
$$
For the bond the terminal condition is $P(T,T)=1$. This argument also shows why stochastic discounting remains inside the conditional expectation.
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