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Let $D_K(S,t)=P_d(S,t)$ and $\gamma =1-2r/{\sigma }^2$. The Black-Scholes reflection symmetry gives another solution $(S/B{)}^{\gamma }{D}_K(B^2/S,t)$. Subtracting it enforces absorption:At $S=B$ the terms agree. At expiry the reflected argument satisfies $B^2/S>B>K$, so its digital-put payoff vanishes; the first term supplies the required terminal payoff. These facts and the bounded pricing solution's uniqueness prove the formula. The up-and-out digital put price is zero once knocked out.

While the up-and-out digital put is alive, its pricing domain is $0<S<B$, $t<T$. Its Black-Scholes equation, terminal financial payoff, and absorbing boundary condition areFor completeness, the natural limiting condition is ${\lim }_{S\downarrow 0}H(S,t)=e^{-r(T-t)}$, with $0\le H(S,t)\le e^{-r(T-t)}$. A positive geometric Brownian motion cannot hit zero in finite time, and from a very small starting value the survival and below-strike terminal events have probability tending to one. If the barrier has already been hit, its value is zero regardless of the current stock price.

For $K>B$, the terminal payoff tends to one from inside at $S=B$, while an already hit barrier pays zero. Thus there is a terminal-boundary corner discontinuity; the terminal condition is imposed only for $S<B$, not as a requirement of continuous extension at $(B,T)$.

If $0<B<K$, every surviving terminal price is already below the strike, so this is a barrier digital put paying one at expiry exactly when the upper barrier has been avoided. To use the Black-Scholes reflection symmetry, cap the free digital-put strike at $B$, rather than continuing to use $K$:Indeed, $D_B$ has terminal payoff one throughout the alive region, while its reflected argument exceeds $B$ and hence has zero terminal payoff. The barrier values again cancel. Using $D_K$ in both terms would generally leave an unwanted reflected terminal payoff near the barrier.

Writing $b=r-{\sigma }^2/2$ makes the discounted survival probability explicit:The price is independent of $K$ in this case. Both cases combine by replacing the free strike with $\min (K,B)$ in the up-and-out digital put formula.