The risk neutral probability is \(q=\frac{1+R-d}{u-d}= \frac12\). The price of the option is given by
\[C(0)=P_*\left(\Gamma\right),\] where \(\Gamma\) is the event that for every \(k\), the inequality \(S(k)\leq B_k\) holds. Let \(Z_i\) be the factor by which the price is multiplied at the period \(i\). Then we have \(S(i)=S(0)\cdot Z_1\cdot Z_2\cdots Z_i\), and if we denote \(X_i=\ln { Z_i}\), then we have \(S(i)=S(0)e^{X_1+\cdots+X_i}\) and \(X_i\in\{\ln u,\ln d\}\). Let \(W_i=X_1+\cdots+X_i\). The inequality \(S(k) < B_k\) can be rewritten as
\[W_k\leq k\cdot \frac{\ln u+\ln d}2.\] Notice that \(E_*(X_1)=\frac{\ln u+\ln d}2\). Let us denote \(\mu=E_*(X_1)\).
Therefore \begin{eqnarray*}C(0)&=&P_*\left(\Gamma\right)=P_*\left(X_1+X_2+\cdots+X_k\leq \frac k2\left(\ln u+\ln d\right)\mbox{ for every }k\right)
\newline
&=&P_*\left((X_1-\mu)+(X_2-\mu)+\cdots+(X_k-\mu)\leq 0\mbox{ for every }k\right).
\end{eqnarray*}
Let \(Y_i=\frac{X_i-\mu}{\frac{\ln u-\ln d}2}\). Then \(E_*(Y_i)=0\) and \(Y_i\in\{-1,1\}\). We have
\begin{eqnarray*}
C(0)&=&P_*\left(Y_1+\cdots+Y_k\leq 0\mbox{ for every }k\right) = P_*\left(Y_1+\cdots+Y_k< 1\mbox{ for every }k\right)=1-P_*\left(Y_1+\cdots+Y_k\geq 1\mbox{ for some }k\right).
\end{eqnarray*}
We have that \begin{eqnarray*}P_*\left(Y_1+\cdots+Y_k\geq 1\mbox{ for some }k\right)&=&\sum_{j={-98}}^{100}P_*\left(Y_1+\cdots+Y_{100}=j,\;\;Y_1+\cdots+Y_k\geq 1\mbox{ for some }k\right)\newline
&=&
\sum_{j=1}^{100}P_*\left(Y_1+\cdots+Y_{100}=j,\;\;Y_1+\cdots+Y_k\geq 1\mbox{ for some }k\right)\newline && +\sum_{j={-98}}^0P_*\left(Y_1+\cdots+Y_{100}=j,\;\;Y_1+\cdots+Y_k\geq 1\mbox{ for some }k\right)\newline
&=&\sum_{j=1}^{100}P_*\left(Y_1+\cdots+Y_{100}=j\right)+\sum_{j={-98}}^0P_*\left(Y_1+\cdots+Y_{100}=j,\;\;Y_1+\cdots+Y_k\geq 1\mbox{ for some }k\right)\newline
\end{eqnarray*}
According to reflection principle we have that when \(j\in\{-98,-97,\dots, 0\}\) the following holds:
\[P_*\left(Y_1+\cdots+Y_{100}=j,\;\;Y_1+\cdots+Y_k\geq 1\mbox{ for some }k\right)=P_*(Y_1+\cdots+Y_{100}=2-j).\]
Let us denote \(M_n=Y_1+\cdots+Y_{n}\). Then we have
\begin{eqnarray*}P_*\left(Y_1+\cdots+Y_k\geq 1\mbox{ for some }k\right)&=&\sum_{j=1}^{100}P_*\left(Y_1+\cdots+Y_{100}=j\right)+\sum_{j=-98}^0P_*(Y_1+\cdots+Y_{100}=2-j)\newline
&=&P_*\left(M_{100}\geq 1\right)+\sum_{i=0}^{98} P_*\left(M_{100}=i+2\right)\newline
&=&P_*\left(M_{100}\geq 1\right)+P_*\left(M_{100}\geq 2\right)=2 P_*\left(M_{100}\geq 2\right),\end{eqnarray*} because \(P_*\left(M_{100}=1\right)=0\) since the walk cannot end at level \(1\).
Notice that \(P_*(M_{100}\geq 2)=P_*(M_{100}\leq -2)\) and that \(P_*(M_{100}\geq 2)+P_*(M_{100}\leq -2)+P_*(M_{100}=0)=1\), therefore
\(P_*(M_{100}\geq 2)=\frac12\left(1-P_*(M_{100}=0)\right)=\frac12\left(1-\binom{100}{50}\cdot\frac1{2^{100}}\right)\).
Therefore
\[P_*\left(Y_1+\cdots+Y_k\geq 1\mbox{ for some }k\right)=1-\frac{\binom{100}{50}}{2^{100}} .\]
Thus \[C(0)=\frac{\binom{100}{50}}{2^{100}}\approx 0.079589.\]