How to compute $mathbb{P}(B(tau)leq a)$, $B(t)$ is standard Brownian motion.

It boils down to more general result. Namely:

Theorem Let $(Omega,mathcal F,mathbb P)$ be a probability space, $mathcal G subset mathcal F$ a sub-$sigma-$field. Let $X:Omega to (E_1,mathcal E_1)$ and $Y: Omega to (E_2,mathcal E_2)$ be random variables (where $(E_i,mathcal E_i)$ are some let's say polish spaces). Assume that $X$ is independent of $mathcal G$ whereas $Y$ is $mathcal G$ measurable. Moreover let $F:E_1 times E_2 to mathbb R$ be measurable such that $mathbb E[|F(X,Y)|] < infty$. Then $mathbb E[F(X,Y)| mathcal G] = H(Y)$ almost surely, where $H(t) = mathbb E[F(X,t)]$

Proof Take any $A in mathcal G$. We need to prove $$ mathbb E[F(X,Y) 1_A] = mathbb E[H(Y)1_A] $$ To do this, consider random vector $V=(X,Y,1_A)$ and note that due to independence assumptions, $X$ is independent from $(Y,1_A)$, so that letting $mu_X,mu_{(Y,1_A)}$ be respectivelly distributions of $X$ and $(Y,1_A)$ we arrive at $$ mathbb E[F(X,Y)1_A] = int_{E_1 times E_2 times mathbb R} F(x,y)z dmu_{(X,Y,1_A)}(x,y,z) = int_{E_1}int_{E_2 times mathbb R} F(x,y)z dmu_X(x)dmu_{(Y,1_A)}(y,z) = $$ $$ = int_{E_2 times mathbb R} z int_{E_1} F(x,y) dmu_X(x) dmu_{(Y,1_A)}(y,z) = int_{E_2 times mathbb R} z mathbb E[F(X,y)] dmu_{(Y,1_A)}(y,z)$$ Where we used Fubinii due to integrability assumptions. Now note that under integral we have our function $H$, hence we showed $$ mathbb E[F(X,Y)1_A] = int_{E_2 times mathbb R} z H(y) dmu_{(Y,1_A)}(y,z) = mathbb E[1_A H(Y)] $$ which proves our theorem.

In your question, you want to use it here: $$ mathbb E[1_{B(tau) le a} | tau] $$ with $F(B,tau) = 1_{B(tau) le a}$ (we treat here $B$ as a function $Omega to C([0,infty))$ and $mathcal G = sigma(tau)$ getting $$ mathbb P(B(tau) le a) = mathbb E[ mathbb E[F(B,tau) | tau]] = mathbb E[ H(tau)] $$ where $H(t) = mathbb E[F(B,t)] = mathbb P(B(t) le a) = Phi(frac{a}{sqrt{t}})$. Hence $$ mathbb P(B(tau) le a) = mathbb E[ Phi(frac{a}{sqrt{tau}})] = int_0^infty lambda e^{-lambda x} Phi(frac{a}{sqrt{x}})dx $$

Edit: Maybe two words why function $F$ is measurable. Note that $F$ is a superposition of $1_{[-infty,a]}$ which is measurable and function $G:C([0,infty) times mathbb R to mathbb R$ given by $G(f,x) = f(x)$. It is not hard to see that $G$ is continuous, hence measurable. Indeed, taking any such $f,x$ and $varepsilon > 0$ we have for all $g in C([0,infty)$ such that $|f-g|_{infty} le frac{varepsilon}{2}$ and for all $y in mathbb R$ such that $|x-y| < delta$ for appriopate $delta$ giving $|f(x)-f(y)|<frac{varepsilon}{2}$ that $$ |G(f,x)-G(g,y)| = |f(x)-g(y)| le |f(x)-f(y)| + |f(y)-g(y)| le frac{varepsilon}{2} + |f-g|_{infty} le varepsilon $$