Observe that the mean squared error can be expanded as (using linearity of expectation):
$$
\mathrm{MSE}(\theta; T) = \mathbb{E}(T^2) + \mathbb{E}(g(\theta)^2)
-2 \mathbb{E}(T g(\theta))
$$
In your case, $g(\theta)$ is just the population mean $p$, and additionally $\mathbb{E}(T) = p$, so we may rewrite
$$
\mathrm{MSE} = \mathbb{E}(T^2) + p^2 - 2p\mathbb{E}(T) =
\mathbb{E}(T^2) - p^2 = \mathbb{E}(T^2) - [\mathbb{E}(T)]^2
= \mathrm{Var}(T).
$$
It is then straightforward to calculate which of the two estimators has minimum variance, using the fact that $X_i, Y_i$ are all independent.
For example, for the estimator $\frac{\bar{X} + \bar{Y}}{2}$, we get
$$
\mathrm{Var}\left(\frac{\bar{X} + \bar{Y}}{2}\right) =
\mathrm{Var}\left(\frac{1}{2m} \sum_{i=1}^m X_i + \frac{1}{2n}
\sum_{i=1}^n Y_i \right) =
\frac{1}{4m^2} \sum_{i=1}^m \mathrm{Var}(X_i) +
\frac{1}{4n^2} \sum_{i=1}^n \mathrm{Var}(Y_i) \\
= p(1 - p) \left( \frac{1}{4m} + \frac{1}{4n} \right)
$$
where we've used the independence of all $X_i, Y_i$ to interchange variance with summation, the property that $\mathrm{Var}(aX) = a^2 \mathrm{Var}(X)$ when $a$ is a constant, and the fact that all variables are identically distributed as $\mathrm{Bernoulli}(p)$ to pull out the common factor $p(1 - p)$ out of the sum.
Note that if $\hat \theta(X)$ is an estimator (depending on random data $X$) for the parameter $\theta\in \mathbb{R}^n,$ the MSE is a scalar quantity defined as
$$\begin{align}MSE(\hat\theta,\theta)&\equiv E[\|\hat\theta(X)-\theta\|^2]\\
&=E[(\hat\theta(X)-\theta)'(\hat\theta(X)-\theta)].\\\end{align}$$
With some matrix algebra, one can easily prove the identity
$$\begin{align}MSE(\hat\theta,\theta)&=\|Bias(\hat\theta,\theta)\|^2+tr(Var(\hat\theta(X))),\\
Bias(\hat\theta,\theta)&\equiv E[\hat\theta(X)]-\theta. \end{align}$$
So rather than look at a vector of individual MSEs, we typically look at the above metric as the generalization of MSE.
However, the MSE is only one metric to judge an estimator by. One may also be interested in looking at the variance-covariance matrix $Var(\hat\theta(X))$, in which case your question still stands, namely how do we decide which of $V_1\equiv Var(\hat\theta_1(X))$, $V_2\equiv Var(\hat\theta_2(X))$ is "greater" given two estimators $\hat\theta_1(X),\hat\theta_2(X)$?
A common partial order used in this respect that is defined on the set of symmetric positive semidefinite matrices is the Loewner order:
$$V_1\geq V_2\iff V_1-V_2 \text{ is positive semidefinite (p.s.d)}.$$
Being a partial order, this relation cannot be used to compare any two variance-covariance matrices summoned from the ether, but it is still meaningful. For instance, because p.s.d matrices have nonnegative diagonal entries, one immediate implication of $V_1\geq V_2$ is that the variance of each component of $\hat\theta_1(X)$ is at least as great as the variance of the corresponding component of $\hat\theta_2(X).$
Best Answer
EDIT
The two estimator are not the same.
I do not know if the exercise asks you to find analytically the two MSE's, but if $\overline{X}_n\leq\frac{1}{2}$ the two MSE's are the same, and equal to the sample means' variance: $\frac{\theta(1-\theta)}{n}$. On the contrary, if $\overline{X}_n>\frac{1}{2}$ the first estimator does not make sense.
Restricted MLE
in this example, Likelihood's domain is restricted in $\theta \in[0;0.5]$ so it is self evident that if $\overline{X}>0.5$ the likelihood is strictly increasing and its argmax is on the border: $\hat{\theta}_{ML}=0.5$
Let's look at the following example:
Let's draw an unfair coin 10 times. Suppose we have the two following cases
3 Successes on 10 Draws
7 Successed on 10 Draws
the two likelihoods are the following
EDIT2:
Let's have a focus on the MSE(ML)
This changes if the estimator "sample mean" is greater than 0.5 or not.
$$\mathbb{V}[\overline{X}_n]=\frac{1}{n^2}n\mathbb{V}[X_1]=\frac{\theta(1-\theta)}{n}$$