A sketch of the proof is as follows:

Consider the site $Y_{syn-cris}$ where the objects are the same as in $Y_{cris}$ but the
covering families are surjective syntomic families. Then there are maps of topoi:
$\alpha : Sh(Y_{syn-cris})\to Sh(Y_{syn})$ and $\beta : Sh(Y_{syn-cris})\to Sh((Y/W_n)_ {cris})$,
defined by $\beta_*(F)(U,T) = F(U, T)$ and $\alpha_{*}(F)(U) = H^0_{syn-cris}(U/W_n,F)$.

Lemma. $R^i\beta_*\mathcal O_{Y/W_n} = R^i\alpha_*\mathcal O_{Y/W_n} = 0$ for $i > 0$.

This implies the result in your question by a standard application of the Leray spectral
sequence.

As for the lemma: To prove that $R^i\beta_*\mathcal O_{Y/W_n} = 0$ for $i > 0$, it boils
down to checking that if $U$ is an affine in $(Y/W_n)_ {cris}$ then $H^i_{syn}(U,\mathcal O_U) = 0$
for $i > 0$, and this follows the theory of the Cech complex (it is acyclic because of faithful flatness of the syntomic cover).

To prove that $R^i\alpha_*\mathcal O_{Y/W_n} = 0$, we have to check that if $U$ is an open of
$Y_{syn}$, and $s\in H^i_{cris}(U/W_n)$ (strictly speaking we have to do the computation with syntomic-cristalline cohomology, but by the previous part, the cohomology groups coincide with the crystalline cohomology groups) then there exists a syntomic cover $U_i\to U$ such
that $s\mid U_i = 0\in H^i_{cris}(U_i/W_n)$. Now recall that we can compute this cohomology groups as the hypercohomology groups of the de Rham complex of the divided power envelope of some embedding into a smooth scheme. That means, after shrinking, we can represent $s$ as an $i$-form.
We need to find a syntomic cover such that when we restrict $s$ to this cover, it vanishes. To do this, note that $A[T]\to A[T^{p^-1}]$ is a syntomic cover that has the property that the image of
$dT$ is zero.