Given,

Variance $=3$

So,

$\mu_2=3$

Fourth central moment is:

$\mu_4=63$

Coefficient of kurtosis is:

$\beta_2=\frac{\mu_4}{\mu_2^2}$

Substitute the values:

$\beta_2=\frac{63}{3^2}$

$\beta_2=\frac{63}{9}$

$\beta_2=7$

For a normal or mesokurtic distribution,

$\beta_2=3$

Here,

$\beta_2=7>3$

So, the distribution is leptokurtic.Given,

$Z=XY$

Now, $Z=1$ only when both $X=1$ and $Y=1$.

Since $X$ and $Y$ are independent,

$P(Z=1)=P(X=1,Y=1)$

$P(Z=1)=P(X=1)P(Y=1)$

$P(Z=1)=\frac{1}{2}\times \frac{1}{2}$

$P(Z=1)=\frac{1}{4}$

Now,

$P(Z=0)=1-P(Z=1)$

$P(Z=0)=1-\frac{1}{4}$

$P(Z=0)=\frac{3}{4}$

Therefore, $Z$ follows Bernoulli distribution with

$P(Z=1)=\frac{1}{4}$ and $P(Z=0)=\frac{3}{4}$

Correct answer is option $4$.