It is evident that \(\sc{BHP}\) is in \(\textbf{NP}\) since on input \(\langle \ulcorner N \urcorner, x,1^t \rangle\) an efficient universal non-deterministic machine can determine if \(N\) accepts \(x\) in time polynomial in \(\lvert x\rvert\) and \(t\). To see that \(\sc{BHP}\) is hard for \(\textbf{NP}\), observe that if \(Y \in \textbf{NP}\), then \(Y\) corresponds to the set of strings accepted by some non-deterministic machine \(N \in \mathfrak{N}\) with polynomial running time \(p(n)\). If we now define \(f(x) = \langle \ulcorner N \urcorner,x,1^{p(\lvert x\rvert)} \rangle\), then it is easy to see that \(f(x)\) is a polynomial time reduction of \(Y\) to \(\sc{BHP}\).